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A  TREATISE  ON  TRIGONOMETRY. 


TREATISE 


ON 


TRIGOlSrOMETPvY 


BY 


PROFS.  OLIVER,  WAIT  AND  JONES 


CORNELL   UNIVERSITY. 


NEW  YORK: 

JOHN  WILEY  &   SONS. 
1883. 


Entered  according  to  Act  of  Congress,  in  the  year  1881,  by 

GEORGE   WILLIAM  JONES, 
in  the  oflice  of  the  Librarian  of  Congress,  at  Washington. 


PKEFACE. 


THIS  book  is  one  of  a  series  of  text-books  to  be  prepared  by 
the  department  of  mathematics  of  Cornell  University,  in  accord- 
ance with  the  scheme  of  instruction  now  in  force  here.  It  was 
outlined  and  written  mainty  by  Prof.  Jones ;  but  it  has  been 
carefully  read  by  all  of  us,  the  general  plan,  and  all  difficulties, 
have  been  discussed  together,  the  proofs  have  been  submitted 
to  all,  and  it  goes  out  as  our  joint  production.  It  is  designed 
as  a  drill-book  for  class  use  ;  its  leading  features  are  : 

The  general  definition  of  the  trigonometric  functions  in  terms 
applicable  to  all  angles,  without  regard  to  sign  or  magnitude. 

The  expression  of  the  functions  of  all  angles  in  terms  of  the 
functions  of  positive  angles  less  than  a  right  angle,  by  direct 
reference  to  the  definitions. 

The  graphical  representation  of  functions. 

The  general  proof  of  the  formulae  for  the  functions  of  the  sum 
and  difference  of  two  angles,  of  double  angles,  half- angles,  etc. 

The  differentiation  of  trigonometric  functions,  their  develop- 
ment thereby  into  scries,  and  the  computation  of  the  trigono- 
metric canon  by  means  of  these  series. 

The  solution  of  oblique  triangles  by  means  of  right  triangles, 
as  well  as  by  the  general  properties  of  triangles ;  and  by  the 
use  of  natural  as  well  as  logarithmic  functions. 

An  exhaustive  discussion  of  the  ambiguous  and  impossible 
cases  of  right  and  oblique  triangles. 

A  careful  choice  and  arrangement  of  topics,  according  to  their 
relations  to  practical  work  and  to  the  higher  mathematics. 

1 30896 


iv  PREFACE. 

The  exact  statement  of  principles  in  the  form  of  theorems  and 
corollaries,  and  their  rigorous  demonstration. 

Frequent  reference  of  collateral  matter  to  the  reader  for 
demonstration. 

Copious  and  varied  exercises. 

In  the  preparation  of  the  book,  free  use  has  been  made  of  the 
works  of  other  authors,  particularly  those  of  Briot  and  Bouquet, 
De  Morgan,  Todhunter,  Peirce,  Wheeler,  Greenleaf,  Loomis, 
and  Chauveuet;  and  of  collections  of  exercises  representing 
Senate-house  examination  papers. 

The  careful  reader  will  doubtless  find  many  typographical  and 
other  errors  in  this  first  edition ;  he  will  confer  a  great  favor  if 
he  will  kindly  communicate  them  to  either  of  the  authors.  Any 
suggestions  from  practical  teachers,  looking  to  the  improvement 
of  the  book  in  either  matter  or  form,  will  be  welcomed  and 
esteemed  of  great  value. 

Among  other  such  improvements  now  in  contemplation  is  the 
addition  of  a  chapter  on  the  applications  of  spherical  trigonometry 
to  astronomy,  geodesy  and  navigation,  and  one  on  imaginaries, 
and  an  alphabetical  index  to  the  whole. 

The  starred  portions  of  the  book  may  be  omitted  without 
breaking  the  continuity  of  the  subject ;  and  to  such  teachers 
as  do  not  desire  to  take  up  the  whole  treatise  the  following 
abridgment  is  recommended : 

I.    §§  1-23,  except  the  note  to  §  18,  and  Note  4  to  §  19  ;  selec- 
tions from  Ex.  1-7,  9,  21-23,  and  25-28. 
H.    §§1-3. 

III.  §  2  ;    §  3,  one  method,  and   §  1  if  the  second  method  is 

chosen ;  Ex.  1—19  and  26-45. 

IV.  §§  1,   2,   4,   5   and  6,  one  method,  and  §  3  if  the  second 

metJiod  is  chosen,  except  Thms.  8-10  ;  Ex.  1-26. 

O.  W.  J. 

ITHACA,  N.Y.,  April  8,  1881. 


CONTENTS. 

PLANE   TRIGONOMETRY. 
I.    PRIMARY  DEFINITIONS  AND  FORMULAE. 

Page 

§  1.   Plane  angles 

§  2.    Angular  notation 3 

§  3.    Complement  and  supplement  of  an  angle  or  arc 4 

§  4.   Positive  and  negative  lines •  •  5 

§  5.  Position  of  a  point  in  a  plane  by  abscissa  and  ordinate   ...  5 

§  G.   Position  of  a  point  by  bearing  and  distance 7 

§  7.   Trigonometric  functions 

§  8.    Anti-functions •  9 

§  9.    Constancy  of  functions 

§  10.   Periodicity  of  functions 

§11.    Signs  of  functions 10 

§  12.    Functions  of  negative  angles 1° 

§13.    Functions  of  the  complement  of  an  angle 11 

§  14.   Functions  of  .]  T  +  8 12 

§  15.    Functions  of  the  supplement  of  an  angle 12 

Functions  of  TT  +  0 13 

Functions  of  :]-  —  ft 14 

§  18.   Functions  of  ^  +  8 : 

§19.    Values  of  functions  in  terms  of  each  other 15 

§20.    Functions  of  30°,  GO0, ] 

§21.    Functions  of  45°,  135^,  ..... 19 

§22.    Functions  of  0°,  90°, ' 

§  23.   Graphical  representation  of  functions ' 

§  24.   Exercises 


vi  CONTENTS. 

II.    GENERAL  FORMULAE. 

Page. 

§  1.  Functions  of  the  sum,  and  of  the  difference,  of  two  angles  .  .  28 

§2.  Functions  of  double  angles,  and  of  half-angles 31 

§3.  Functions  of  the  half-sum,  and  of  the  half-difference,  of  two 

angles 33 

§  4.  Functions  of  the  sum  of  three  or  more  angles,  of  triple 

angles,  etc 34 

§  5.  Differentiation  of  trigonometric  functions 35 

§6.  Development  of  trigonometric  functions 37 

§7.  The  trigonometric  canon 39 

§  8.  Exercises 44 

III.  SOLUTION  OF  PLANE  TRIANGLES. 

§  1.    General  properties  of  plane  triangles 40 

§2.    Solution  of  right  triangles 51 

§  3.    Solution  of  oblique  triangles 54 

§  4.    The  area  of  a  triangle 59 

§5.   Inscribed,  escribed  and  circumscribed  circles Gl 

§  G.    Exercises     . G2 


SPHERICAL    TRIGONOMETRY. 
IV.    SOLUTION  OF  SPHERICAL  TRIANGLES. 

§  1.    Geometrical  principles G8 

§  2.    Napier's  rules  foe  the  right  triangle 70 

§  3.    General  properties  of  spherical  triangles 73 

§  4.    Solution  of  right  triangles 81 

§5.    Solution  of  quaclrantal  triangles  and  isosceles  triangles  .   .    .  8G 

§6.    Solution  of  oblique  triangles *7 

§  7.   Relations  between  spherical  and  plane  trigonometry    ....  98 

§  8.   Exercises ....  100 


T    '      , 


TRIGONOMETRY. 


TRIGONOMETRY  is  that  branch  of  mathematics  which  treats  of 
the  numerical  relations  of  angles  and  triangles. .  It  is  essentially 
algebraic  in  character,  but  is  founded  on  Geometry.  It  has  two 
parts :  Plane  Trigonometry,  which  treats  of  plane  angles  and 
triangles,  and  Spherical  Trigonometry,  which  treats  of  spherical 
angles  and  triangles. 


PLANE    TRIGONOMETRY. 


I.   PRIMARY  DEFINITIONS  AND  FORMULAE. 

§  1,     PLANE    ANGLES.  • 

A  plane  angle  is  the  opening  between  two  straight  lines  which 
meet  at  a  point,  and  is  generated  b}'  revolving  one  of  the  lines 
about  the  common  point  as  a  hinge.  The  fixed  line  is  the  initial 
line,  and  the  moving  line  is  the  terminal  line.  The  revolving 
line  has  made  one  revolution  when,  by  a  continuous  forward 
motion,  it  has  come  again  to  the  position  it  first  occupied.  An 
angle  is  not  limited  in  Trigonometry  by  the  geometrical  words 
"  acute"  and  ''obtuse"  ;  it  may  exceed  a  half  revolution,  or  a 
whole  revolution,  or  it  may  be  any  number  of  revolutions,  thus : 


PLANE   TRIGONOMETRY. 


[I. 


If  the  line  OP  first  coincide  with  ox,  and  then,  revolving  about 
o,  take  successively -the  positions  opj,  op2, 

OPS,  OP4,  OFj,  ,  the  angles  generated  are 

XOPI?  xop2,  XOPS.  xop4, ,  and  these  angles, 

by  general  agreement,  are  called  positive 
angles, 

while  xop4,  XOPS, are  negative  angles. 

That  is  to  say,  revolution  from  right  to  left, 
and  opposite  to  the  hands  of  a  clock,  is 

positive  revolution;  and  revolution  from  left  to  right  is  negative 

revolution.     The  order  of  the  letters  indicates  whether  the  angle 

is  positive  or  negative,  thus  : 

xopj  is  positive,  and  Pjox  is  negative ; 
xop14-PiOx  =  0, 

4-  PI  op2  4-  P2  ox  =  0, 


but       PiOp2  4-  P2Op3  4-  P3op4  4~  P4°pi  =  owe  entire  positive  revo- 
lution, 

and      PiOP4  4-  P4OP3  -f-  P3OP2  -f-  P2OP!  =  one  entire  negative  revo- 
lution. 

So,  the  arcs  generated  by  a  point  p,  moving  along  a  circle,  are 
positive  or  negative  arcs,  according  as  the  movement  of  the  point 
indicates  positive  or  negative  revolution  of  the  radius  OP,  and 
therefore,  according  as  the  angles  which  they  subtend  are  posi- 
tive or  negative  angles,  thus  : 

XP15  XP2,  are  positive  arcs, 

but       xp4,  xp3,  are  negative  arcs; 

and      xp14-PiP24-P2P3  +  P3P24-P2X  =  0, 
but       Xp1  +  P1p2  +  P2p34-P3p44-P4x  =  one 
circumference. 

If  Y'Y  J_x'x  at  o  ;  then  :  _^ 

XOY  is  the  first  quadrant,  x' 

YOX'  is  the  second  quadrant, 
X'OY'  is  the  third  quadrant, 
Y'OX  is  the  fourth  quadrant. 


§  2.]  PRIMARY  DEFINITIONS   AND  FORMULAE.  3 

Aii  angle  is  said  to  be  in  the  first,  second,  third,  or  fourth 
quadrant,  according  as  its  terminal  line  lies  in  the  first,  second, 
third,  or  fourth  quadrant,  counting  from  the  initial  line. 

If  the  revolution  of  the  terminal  line  is  continuous,  it  comes 
again  and  again  to  the  same  positions  which  it  had  during  the 
first  revolution  ;  such  returns  are  periodic. 

§  2,     ANGULAR    NOTATION. 

The  right  angle  is  divided  into  90  equal  parts  called  degrees, 
the  degree,  into  GO  equal  parts  called  minutes;  the  minute,  into 
60  equal  parts  called  seconds.     Degrees,  minutes  and  seconds 
are  marked  °,  ',  ",  respectively,  thus : 
6°  29'  33". 3  is  read  G  degrees,  29  minutes,  33  and  3  tenths  seconds. 

When  angles  are  expressed  in  degrees,  minutes  and  seconds, 
they  are  said  to  be  expressed  in  degree-measure. 

Another  measure  called  circular  measure,  arcual  measure,  or, 
briefl}',  -rr-measure,  comes  from  the  use  of  the  ratio  of  an  arc 
which  subtends  an  angle,  to  the  radius  of  the  arc ; 
for,  since,  in  the  same  or  equal  circles,  angles  at  the  centre  are 
proportional  to  the  arcs  which  subtend  them,  and,  in  different 
circles,  like  arcs  are  proportional  to  their  radii,  [geom. 

therefore  the  ratio,  arc :  radius,  is  proportional  to  the  angle  sub- 
tended, and  may  be  used  as  a  representative  or  measure  of  it. 

But  the  ratio,  half-circle  :  radius,  equals  3.14159 ,  called  TT  ; 

therefore  a  half-revolution,  or  an  angle  of  180°,  is  represented 

by  TT  ;  a  right  anglo,  or  90°,  by  ^ ;  a  whole  revolution,  or  360°, 

by  277  ;  30%  by  -  ;  and  so  on. 

G 

That  angle  whose  arc  is  as  long  as  the  radius,  and  which  may 

>  1  ono 

be  called  the  unit  angle  of  the  7r-measure,  is  -— ,  =  57°  17'  44".8. 

7T 

The  7r-measure  of  an  angle  of  1°  is  — ,  ==  .017  4533  ; 

180 

that  of  1'  is  .017  4533  -j-  GO,  =  .000  2909  ; 
that  of  I"  is  .000  2909  --  GO,  =  .000  0048. 

An  angle  given  in  7r-measure  may  be  expressed  in  degree- 
measure  by  multiplying  it  into  the  unit  angle  of  7r-measure  ; 


4  PLANE   TEIGONOMETKY.  [I. 

and  an  angle  given  in  degree-measure  may  be  expressed  in 
TT-measure  by  multiplying  the  number  of  degrees,  minutes  or 
seconds  by  the  --measure  of  1°,  1'  or  1". 

NOTE.  The  reader  must  carefully  distinguish  between  the  two 
notations,  for  though  90°  and  ^  both  represent  a  right  angle,  it 
is  hardly  right  to  say  90°  =  -,  for  90°  is  a  right  angle,  an  actual 
geometrical  magnitude,  while  -  is  merely  a  ratio,  i.e.,  a  number. 

With  this  caution,  however,  the  angle,  the  arc,  and  the  ratio, 
arc:  radius,  may  be  used  almost  at  pleasure,  and  the  one  nota- 
tion or  the  other  may  be  employed  at  convenience.  The  7r-meas- 
ure  is  generally  preferable  for  theoretical  work,  but  the  degree- 
measure  for  computation  of  triangles. 

§  3,    COMPLEMENT  AND  SUPPLEMENT  OF  AN  ANGLE  OE  ARC. 

The  complement  of  an  angle  is  its  defect  from  a  right  angle ; 
of  an  arc,  its  defect  from  a  quadrant;  i.e.,  it  is  the  remainder 
when,  from  a  right  angle  or  quadrant,  the  given  angle  or  arc  is 
subtracted. 

The  supplement  of  an  angle  is  its  defect  from  two  right  angles  ; 
of  an  arc,  its  defect  from  a  half-circle ;  i.e.,  it  is  the  remainder 
when,  from  two  right  angles  or  a  half-circle,  the  given  angle  or 
arc  is  subtracted. 

Manifestly  the  complement  of  a  positive  angle  or  arc  less  than 
90°  is  a  positive  angle  or  arc  less  than  90°  ;  of  a  positive  angle 
or  arc  greater  than  90°,  is  a  negative  angle  or  arc  ;  of  a  negative 
angle  or  arc,  is  a  positive  angle  or  arc  greater  than  90°. 

So,  the  supplement  of  a  positive  angle  or  arc  less  than  180°  is 
a  positive  angle  or  arc  less  tha,n  180°  ;  of  a  positive  angle  or  arc 
greater  than  180°,  is  a  negative  angle  or  arc  ;  of  a  negative  angle, 
or  arc,  is  a  positive  angle  or  arc  greater  than  180°.     Thus  : 
the  complement  of  75°  is  15° ;  of  100°,  is  — 10° ;  of  — 10°,  is  100° ; 

of -,  is--;  of27r,  is'-—  i 
0         o  2 

the  supplement  of  75°  is  105°;  of  200°,  is  —20°;  of  —20°,  is 

200°  ;  of  -,  is  —  ;  of  2  TT,  is  -TT. 
6         6 


§5.]  PRIMARY  DEFINITIONS  AND  FORMULAE. 


§  4,     POSITIVE   AND   NEGATIVE   LINES. 

If  two  or  more  points  lie  on  a  straight  line,  and  the  position 
of  any  one  of  them  be  known,  then  the  positions  of  the  other 
points  are  determined  b}'  their  distances  and  directions  from  the 
point  first  named.  The  first  point  is  the  oriyin;  and,  of  the  two 
directions  from  this  point,  along  the  line,  if  either  be  called  posi- 
tive, the  other  is  negative,  and  the  segments  of  the  line  measured 
in  one  direction  are  positive,  in  the  other  negative.  The  order 
of  the  letters  at  the  extremities  of  a  segment  indicates  the  direc- 
tion of  the  segment,  thus  : 

if  A  and  B  be  two  points  on  a  line,  and  AB  be  positive, 
then  BA  is  negative,  and  vice  versa;  and,  in  either  case, 


G  F  E  D 

So,  if  A,  B,  c  be  any  three  points  on  a  line,  then,  whatever 
their  order  upon  the  line, 

AB  -f  BC  =  AC, 
and      AB  +  BC-f-CA  =  0. 

So,  if  A,  B,  c,  D, K  be  n  points  on  a  line,  then,  whatever 

their  order  upon  the  line, 

AB  -f-  BC  -f  CD  -f =  AK, 

and      AB  -f  BC  +  CD  -{-.....  -J-KA  =  0. 

§  5,     POSITION   OF  A  POINT  IN  A  PLANE  BY  ABSCISSA  AND 

ORDLNATE. 

Let  p  be  an}'  point  in  a  plane,  o  a  fixed  point,  and  ox  a  fixed 
line  in  the  plane  ;  draw  AP  J_  ox  ;  then  the 
position  of  P  with  reference  to  o  and  ox  is 
determined  when  the  length  and  directions 
of  OA  and  AP  are  known.     In  this  figure  : 

the  reference-point  o  is  the  origin;  x  AS  AlX 

the  horizontal  distance  OA  is  the  abscissa, 
and  the  perpendicular  AP  is  the  ordinate,  of  P. 


X'A 


6  PLANE  TEIGOXOMETRY.  [I. 

The  line  x'ox  is  the  axis  of  abscissas, 

and          Y'OY  is  the  axis  of  ordinates. 

The  abscissa  and  ordinate  of  a  point,  when  spoken  of  together, 
are  its  coordinates. 

By  general  agreement,  when  the  line  ox 
is  so  placed  before  the  reader  that  x  is  its 
right  extremit}',  then  ox  is  assumed  as  tlie 
positive  direction  of  the  axis,  and  ox'  as  its 
negative  direction ; 

therefore  abscissas  of  points  in  the  first  and 
fourth  quadrants  are  positive,  and  abscissas 

of  points  in  the  second  and  third  quadrants  are  negative. 

By  general  agreement,  also,  that  side  of  the  axis  from  which 

positive  revolution  is  had  is  assumed  to  be  positive,  and  that  side 

from  which  negative  revolution  is  had  is  negative  ; 

therefore  ordinates  of  points  in  the  first  and  second  quadrants 

are  positive,  and  ordinates  of  points   in  the  third  and  fourth 

quadrants  are  negative. 

The  signs  of  abscissa  and  ordinate  when  written  together  are  : 

respectively,  for  points  in  the  four  quadrants  taken  in  their  order. 
For  brevity,  the  abscissa  of  a  point  may  be  denoted  b.y  x  and 
the  ordinate  by  ?/,  and  the  position  of  a  point  is  then  determined 
from  the  equations 

x  =  a      and      y  =  6, 

wherein  a,  with  its  sign,  stands  for  the  length  and  direction  of 

the  abscissa  of  the  point, 

and  &,  with  its  sign,  for  the  length  and  direction  of  its  ordinate. 

If  x  =  0,  the  point  is  on  the  axis  of  ordinates  ; 
if  y  =  0,  it  is  on  the  axis  of  abscissas  ; 
if  both  x  and  y  =  0,  it  is  at  the  origin. 

If  r  represents  the  distance  between  two  points,  PA  and  P2 ; 
then,  the  difference  of  then*  abscissas,  x^—  x2, 
and  the  difference  of  their  ordinates,  y\—yi, 
are  the  projections  of?*  upon  the  two  axes,  respectively  ; 
and       r  =  (xl  —  x.2)  -  -f-  ( ijl  —  y.2) 2 . 


§  6.]  PRIMARY  DEFINITIONS   AND   FORMULAE.  7 

§  6.     POSITION  OF  A  POINT  BY  BEAEING  AND  DISTANCE. 

When  two  straight  lines  meet,  the  point  of  intersection  is  the 
origin  ;  one  of  the  lines  is  the  initial  line,  the  other  is  the  termi- 
nal line,  and  the  angle  between  them  is  the  bearing,  from  the 
initial  line,  of  any  point  on  the  terminal  line. 
11',  then,  the  distance  of  any  point  from  the  origin,  and  its  bear- 
ing from  the  initial  line,  be  known,  the  position  of  the  point  is 
determined. 

The  positive  directions  of  the  lines  which  enclose  an  angle, 
taken  with  reference  to  that  angle,  are  from  the  origin,  i.e.,  from 
the  vertex  of  the  angle,  measured  along  the  sides  of  the  angle ; 
thus,  with  reference  to  XOP, 
the  sides  ox  and  OP  are  positive, 
and          ox'  and  OP'  are  negative. 
But,  with  reference  to  X'OP', 
the  sides  ox'  and  OP'  are  positive, 
and  ox  and  OP  are  negative. 

So,  with  reference  to  POX'  or  to  X'OP, 
the  sides  OP  and  ox'  are  positive, 
and  OP'  and  ox  are  negative. 

Since  the  terminal  line  may  revolve  to  the  right  or  to  the  left, 
and  since  the  distance  may  be  measured  in  either  the  positive  or 
the  negative  direction,  it  is  manifest  that  a  point  may  be  deter- 
mined by  four  different  combinations  of  bear- 
ing and  distance. 

Let  r  stand  for  the  distance  and  0  for  the 

'^\'Q=-iif 
bearing  of  the  point  p  ;  then  either :  ,?/'  ,\  V* 

r  =  +OP,     and  6  =  +XOP  ; 

>  « 

or         r  =  ~OP,     and  0  =  +XOP';  V'  *f 

or         r  =  "OP,     and  0  =  ~XOP'; 

wherein  the  signs  +  and  ~  are  signs  of  quality,  and  not  of  opera- 
tion ;  they  show  whether  the  line  OP  and  the  angle  XOP  are  taken 
in  the  positive  or  the  negative  direction.  For  the  purposes  of 
this  treatise  the  first  two  combinations  are  sufficient. 


or         r  =  +OP,     and  6  =  ~XOP  ;  *  x- — \       ,-V 


PLANE   TRIGONOMETRY. 


[I- 


§  7,     TRIGONOMETRIC   FUNCTIONS. 


X 


AX 


Let  XOP  be  any  angle,  positive  or  negative,  ox  the  initial  line, 
OP  the  terminal  line,  and  p  any  point  upon  it.     Draw  AP_!_OX  ; 
then  OP  is  the  distance  of  the  point  p,  OA  is  its  abscissa,  AP  is  its 
ordinate,  and  the  six  ratios  between  the  three  lines  r,  x,  and  y 
are  the  trigonometric  f auctions  of  the  angle  XOP,  viz. : 


The  ratio 

i.e., 

is  the 

written 

ordinate  :  distance 
abscissa  :  distance 

yir 
x:  r 

sine  of         XOP 
cosine  of      XOP 

sin  XOP 

COS  XOP 

ordinate  :  abscissa 
abscissa  :  ordinate 
distance  :  abscissa 

y:x 
x:y 
r:  x 

tangent  of    XOP 
cotangent  of  XOP 
secant  of      XOP 

tan  XOP 
cot  XOP 
sec  XOP 

distance  :  ordinate 

r:y 

cosecant  of  XOP 

CSC  XOP 

Two  subsidiary  functions,  sometimes  used,  are  the  versed  sine 
and  coversed  sine;  their  definitions  are  : 

vers  XOP  =  1  —  cos  XOP,     covers  XOP  =  1  —  sin  XOP. 
NOTE.   From  these  definitions  result  directly  the  six  equations  : 
ordinate  =  distance .  sin  XOP  ;     distance  =  ordinate .  esc  XOP  ; 
abscissa  =  distance .  cos  XOP  ;     distance  =  abscissa .  sec  XOP  ; 
ordinate  =  abscissa .  tan  XOP  ;     abscissa  =  ordinate .  cot  XOP. 
The  sine  and  cosine  may  thus  be  called  projecting  factors,  since 
their  effect  as  multipliers  is  to  project  the  distance  upon  the  axes 
of  ordinates  and  of  abscissas  respectively.     So,  the  tangent  and 
cotangent  may  be  called  interchanging  factors,  since  their  effect 
is  to  convert  abscissas  into  ordinates,  and  vice  versa. 


§  10.]          PRIMARY   DEFINITIONS   AND   FORMULAE.  9 

§  8,    ANTI-FUNCTIONS. 

The  expressions  sin"1^,  cos-1a,  tan~]a,  are  called  anti- 
functions,  and  are  read  the  anti-sine  of  a,  the  anti-cosine  of  a, 

They  moan  "the  angle  whose  sine  is  a,"  "  the  angle  whose 

cosine  is  a,"  and  so  on  ;  thus  : 

if          a  =  sin0,     then    0  =  sin"1a; 

if          &  =  cos0,    then    0  =  cos~16;  

§  9,     CONSTANCY  OF  FUNCTIONS. 

THEOREM  1.  The  functions  of  a  given  angle  are  constant, 
ivhatever  points  P',  P",  are  taken  on  the  terminal  line. 

For,  from  P',  P",  draw  A'P',  A"P", JLox;  then 

the  triangles  OA'P',  OA"P", are  similar, 

and  the  ratios  of  their  homologous  sides  are  ,  / 

,      .  p/ 

equal,  viz. : 

y':  r'  =  ?/":?•",  = 

y' :  x'  =  y"  :  #",  = Q.E.  D.  \_geom. 

§  10,     PERIODICITY   OF   FUNCTIONS. 

THM.  2.  The  functions  of  any  angle  0  and  o/2n?r+0  (n  being 
any  integer,  positive  or  negative)  are  identical ;  sine  with  sine, 
cosine  icith  cosine,  and  so  on. 

For,  '.-  ±2?r,  ±4?7, ± 2 me  stand  for  one,  two, n  entire 

revolutions,  forward  or  backward, 

.  • .  OP  has  the  same  position  for  the  angles  0,  ±  2  77  +  0, 

±477+0,  ±2  7177  +  0;  and  r,  x  and  y  are  identical,  each 

with  each,  for  the  angles  thus  formed. 

.*.  the  ratios  arc  identical ;  sine  with  sine,  cosine  with  cosine, 
and  so  on.  Q.  E.  D. 

COROLLARY  1.  The  functions  of  any  negative  angle,  —  0,  and 
of  the  positive  angle,  2  77—  0,  which  has  the  same  bounding  lines, 
arc  identical. 


10 


PLANE  TRIGONOMETRY. 


[I- 


NOTE.  The  trigonometric  functions  are  therefore  periodic 
functions  of  angles,  and  Trigonometrj*  is  sometimes  defined  as 
4  '  that  branch  of  Algebra  which  treats  of  periodic  functions."  If 
6  =  sin'Vt,  then  6  is  airy  one  of  an  infinite  number  of  angles  all 
of  which  have  the  same  sine,  a. 


§  11,     SIGNS   OF   FUNCTIONS. 

THM.  3.     Tlie  signs  of  functions  of  angles  in  the  four  quad- 
rants are  : 


Angle  in 

sin  and  esc 

cos  and  sec 

tan  and  cot 

vers  and 
covers 

First  quadrant 
Second  quadrant 
Third  quadrant 
Fourth  quadrant 

: 

-f 

; 

J 

For,  •••  r  is  measured  along  the  side  of  the  angle,  and  is 
therefore  always  posijtive  ;  [§  6 

and  •••  x  is  positive  in  the  first  and  fourth  quadrants,  and 
negative  in  the  second  and  third  ;  [§  5 

and  •••  y  is  positive  in  the  first  and  second  quadrants,  and 
negative  in  the  third  and  fourth  ;  [§  5 

.*.  the  signs  of  the  several  ratios  are  as  given  above. 

§  12,     FUNCTIONS   OF   NEGATIVE   ANGLES. 

THM.  4.  For  any  negative  angle,  —  0,  the  values  of  the  several 
functions,  in  terms  of  the  functions  of  the  opposite  positive  angle, 
+  0,  are: 

1]  sin  (— 0)  =  — sin0;  esc  (— 0)  =  - csc0  ; 

2]          cos  (—  0)  =  -fcos0;  sec  (—  <9)  =  -f-sec0; 

3]          tan(— 0)  =  — tan0;  cot  (—  0)  =  —  cot0  ; 

For,  let  xopj  be  any  negative  angle,  —  0,  and  about  the  initial 
line  ox,  as  an  axis  of  symmetry,  draw  OP2,  making  XOPS  oppo- 
site to  xoPi ;  i.e.,  of  equal  magnitude,  but  of  opposite  sign. 


§13.J 


PRIMARY  DEFINITIONS   AND  FORMULAE. 


11 


Take  OPi  and  OP^  equal  distances  on  the  terminal  lines,  and  join  PJ  p2; 


then,  since  ox  bisects 
? 


at  right  angles, 
05 


[geom. 


therefore  the  abscissas,  xt  and  x.2,  of  P!  and  P2  are  identical, 
and  their  ordiuates  yl  and  y2  are  opposites  ; 

2/1  _  _  fa  X_L_%2 

»i~~      'a'      ?*i~»V 


/y» 


Q.  E.  D, 


§  13,  FUNCTIONS  OF  $TT— 6,  THE  COMPLEMENT  OF  AN  ANGLE  0. 

TI-IM.  5.     Any  function  of  the  complement  of  an  angle,  co-0  or 
I-TT— 0,  is  the  co-function  of  the  angle,  i.e. : 


4] 
5] 
6] 
\Y 


sin  co-0  =  cos  6  ; 
cos  co-0  =  sin  0  ; 
tan  co-0  =  cot0; 


csc  co-0  =  sec  0 ; 
sec  co-0  =  csc  0 ; 
cot  co-0  =  tan  0. 


AX   A  o  X     i         4>^0  X 

!    ^^  V"'.XX^ 

p^  \JP 

For,  let  XOP  be  any  angle  0,  and  POY  its  complement,  £TT— 0. 
Draw  AP,  ordinate  of  P  with  reference  to  ox  ;  take  OY  =  OP,  and 
draw  BY,  ordinate  of  Y  with  reference  to  OP  ;  then, 

AYBO  =  AOAP, 

OB  =  AP,  and  BY  =  OA  ;  and  OY  =  OP  ; 
i.e.,    x'  =y,       y'  =  %,   and  r'  —  r. 

sin  POY  =?/':?•'  =  x :  r  =  cos  XOP  ; 

cos  POY  =  x'  :r'  =  y:r  =  sin  XOP  ; 
and  so  for  the  other  functions,  as  the  reader  may  prove. 


\_geom. 
\constr. 


12  PLANE   TEIGONOMETHY.  [I. 

NOTE.  The  words  cosine,  cotangent,  and  cosecant  are  ab- 
breviated forms  for  complement-sine,  complement-tangent,  and 
complement-secant  ;  i.e.,  for  sine  of  complement,  tangent  of  com- 
plement, and  secant  of  complement. 


§  14,    FUNCTIONS  OF 

THM.  6.    For  any  angle,  -J-7T-J-0,  the  values  of  the  several  func- 
tions, in  terms  of  the  functions  of  0,  are  : 

sin  (4-7T  +  0)  =  +  cos#  ;         CSC^TT  -f  6)  =  -f  sec<9  ; 
COS(£TT  +  0)  =  —  sin  (9  ;         sec^Tr  -f  6)  =  —  Csc0  ; 
+  0)  =  -  cot  0  ;         cot(^7r  -f  0)  =  —  tantf. 


B_  j  ij 

AX  I     A>'0 


For,  let  XOP  be  any  angle  6,  and  POQ  =  \ir  ;  then  XOQ  =  $7r+0. 
Take  OQ  =  OP,  and  draw  AP  and  BQ  ordinates  of  p  and  Q  with 
reference  to  ox  ;  then, 

•••     AQBO  =  AOAP,  \_geom. 

.•.  OB  =  —  AP,  and  BQ  =  OA  ;  and  OQ  =  OP  ;  [constr. 

i.e.,  x'  =  —  y,  y'  =  x,  and  r'  =  r. 


.'.     sin  XOQ  =  y'  :  r'  =     x:r=      cos  XOP  ; 
cos  XOQ  =  x'  :  r'  =  —  y  :  r  =  —  sin  XOP  ; 
and  so  on,  as  the  reader  may  prove. 

§  15,   FUNCTIONS  OF  TT—  (9,  THE  SUPPLEMENT  OF  AN  ANGLE  6. 

THM.  7.   For  the  supplement  of  any  angle,  ir—Q,  the  values  of 
the  several  functions,  in  terms  of  the  functions  ofO,  are: 
10]       sin  O-  0)  =  +  sin  0  ;         CSC(TT—  0)  =  +  csctf  ; 
11]       COS(TT—  (9)  =  —  cos0;         sec(?r—  #)  =  —  sec(9; 
12]       tan(7r-  0)  =  -  tan0  ;         cot(ir-  0)  =  -  cote. 


§16.] 


PRIMARY   DEFINITIONS    AND   FORMULAE. 


13 


For,  let  XOP  be  any  angle  6,  and  draw  OQ  so  that  /  QOX'=XOP  ; 
then  ZxoQ=7r— 0. 


/•• 


X'"*A 

Take  OQ  =  OP,  and  dratv  AP  and  BQ,  ordinates  of  P  and  Q  with 
reference  to  ox  ;  then, 

•.•     AOBQ  =  AOAP,  \_geom. 

.-.     OB  =  —  OA,     and  BQ  =  AP  ;     and  OQ  =  OP  ;  [constr. 

i.e.,      x'  =  —  x,  y'  =  y'>        an(l    r'  =  r. 

.-.     sin  XOQ  =?/':?•'=      y  :  r  =      sinxop  ; 
COSXOQ  =  x' :  r'  =  —  x:r  =  —  COSXOP  ; 
and  so  on,  as  the  reader  may  prove. 

§  16,    FUNCTIONS  OF  TT  +  B. 

THM.  8.     For  any  angle,  n+O,  the  values  of  the  several  func- 
tions, in  terms  of  the  functions  ofO,  are: 
13]       sin(7r  +  #)  =  —  sin0;         csc(7r  +  (9)  =  — 
14]       cos(7r-f-<9)  =  —  cos0; 
15] 


For,  let  XOP  be  anj*  angle  B,  and  produce  PO  ;  then  XOQ  =  -rr+O. 
Take  OQ  =  OP,  and  draw  AP  and  BQ  ordinates  of  p  and  Q  with 
reference  to  ox  ;  then, 

v  AOBQ  =  AOAP,  \_geom. 

.-.  OB  =  —  OA,     and  BQ  =  —  AP  ;     and  OQ  =  OP  ;        [constr. 

i.e.,  x1  =  —  x,  y'  =  —  y,        and    r' =  r. 

.'.  etc.,  as  the  reader  may  prove. 


14  PLANE  TRIGONOMETRY.  [I. 

o 

§  17,     FUNCTIONS  OF  —  --  0. 

q 

THM.  9.    For  any  angle,  *-j-  —  $,  the  values  of  the  several  func- 
tions, in  terms  of  the  functions  ofO,  are: 

16]       sin  (^-^-costf;         cscf?f  _  0\  = 

' 
17]       c 


\2 

18]       tan^-^=  +  cot^;         cot^5f 
\  z  '      /  \  2 

The  reader  may  prove. 

§  18,    FUNCTIONS  OF 


THM.  10.   For  any  angle,  ^+0,  ^e  voZites  of  the  several  func- 
tions, in  terms  of  the  functions  of  6,  are: 

19]       sin^  +  A^-costf;         cscf—  +      =  -  sec0  ; 
\  2         /  \  2 


20]      cos       T  _|_      =  +  sill  $  .        sec     5  +      =  +  CSC0  ; 
\  *        J  \  *        / 

21]       tan         +      =-cot0;         cot—  +0    =  -  tan0. 


The  reader  may  prove. 

*  NOTE.     The  ten  theorems  just  proved  may  be  summarized 
and  generalized  as  follows  : 


m 


§19.] 


PlllMARY  DEFINITIONS   AND   FORMULAE. 


15 


The}'  also  give  the  following  : 

25]  sin-1(sm0)  =  csc-1(csc0)=2?i7r+0     or    (271+1)^-0; 
26]  cos"1  (cos 0)  =  sec"1  (sec  0)  =  2mr±0 ; 
27]   tim--l(tsLn6)  =  CQt-l(cotO)=nff+0. 

§  19,     VALUES  OF  FUNCTIONS  IN  TEEMS  OF  EACH  OTHER. 

THM.  11.    For  any  angle  6,  the  values  of  the  several  functions, 
in  terms  of  each  other,  are : 


[28] 
Binds 

[29] 
cos0  = 

[30] 
tan0  = 

[31] 

C0t0  = 

[32) 
sec0  = 

[33] 

CSC0  = 

sin0 

VI      sin20 

sin0 

Vl—  sin-0 

1 

1 

VI     cos20 

• 

COS0 

Vl-siir0 
Vl-cos20 

sin0 

COS0 

Vl  —  siu-0 

1 

siu0 
1 

COS0 

Vl  —  eos^ 

COS0 

Vl-cos20 

tan0 

1 

t'in  A 

1 

\/t111-^J    1 

Vtair'0-j-l 

Vtair'0  +  1 
1 

Vtair^  +  1 
cot^ 

1 

tan0 

/%/-\f  /3 

VcoW+I 

tan0 

Vcot20-|-l 
Vsecr'0—  1 

Vcot-^+1 
1; 

C0t0 

COtt/ 

1 

C0t0 

VCOt  U-\-L 

sec0 

sec0 
1 

sec^ 
Vcsc'-'^  —  1 

"   1 

Vsec'20-l 

sect/ 

CSC0 

Vsec20—  1 

csc0 

csc^ 

Vcscr0—  1 

Vcsc20-l 

Vcsc20-l 

CSC0 

cos0  —  x:r, 

csc0  =  r  :  ?/,  sec0  =  r  :  a;, 

=  l,      cos0-sec0  = 


For  •.•  sin  6  —  ?/  :  r, 
and 

34] 

35] 
i.e., 

and 


cot  0  =  x  :  y,     [§  7 
tan0-cot0  =  l; 


cos0  sin  0 

sine  and  cosecant  are  reciprocals, 
cosine  and  secant  are  reciprocals, 
tangent  and  cotangent  are  reciprocals  ; 


16  PLANE  TRIGONOMETRY.  [I. 

whatever  value  is  found  for  sin  (9,  its  reciprocal  is  a  value  of  csc#, 
whatever  value  is  found  for  cos  (9,  its  reciprocal  is  a  value  of  sec#, 
whatever  value  is  found  for  tan  0,  its  reciprocal  is  a  value  of  cot  0, 
and  whatever  values  are  found  for  sin  6  and  cos  0,  their  ratios  are 
values  of  tan  0  and  cot#. 
So,  •'.•  3?  -f  y2  =  r2,  [geom. 


36]  i.e.,   sin2 

=  Vl  —  cos26>,      and  cos 0=  Vl—  siu2(9; 


and  (2)    !+£=£, 

37]  i.e.,   1-f  tan2<9=sec20; 


tan<9  =  Vsec2#— 1,     andsec#  =  Vl  +  tan2< 

38]  i.e., 


and  (3)    ^  +  1=^.; 
y-  y* 


1=Vcsc26'— 1,     andcsc<9  =  Vl+cot2^. 

So,  by  combination  of  the  formulae  which  have  been  proved, 
others  are  established  ;  for  example  : 

sec0      Vl+tanV 
and  • 


cos^ 

sin  0  =  tan  0.  cos  (9=       tan^ 


Vl+tan2(9 

NOTE  1.     These  formulae,  when  taken  two  and  two,  are 
metrical,  for  example : 

(1)  Those  for  sine,  in  terms  of  cosine,  tangent,  secant,  

with  those  for  cosine,  in  terms  of  sine,  cotangent,  cosecant,  

(2)  Those  for  tangent,  in  terms  of  sine,,  cosine,  secant,  

with  those  for  cotangent,  in  terms  of  cosine,  sine,  cosecant,  

(3)  Those  for  secant,  in  terms  of  sine,  cosine,  tangent,  

with  those  for  cosecant,  in  terms  of  cosine,  sine,  cotangent,  


§  19.]          PRIMARY  DEFINITIONS   AND  FORMULAE.  17 


NOTE  2.     For  a  given  ^  ^  the-<  has  but  one 

value  ; 
for  a  given  <  the-!        «    has  but  one  valae  ; 


for  a  given  ^  *H  has  but  one  value  ; 


but  in  every  other  case  there  are  two  corresponding  values,  the 

one  positive,  and  the  other  negative,  for  every  function. 

This  appears  alike  from  the  double  signs  of  the  radicals  involved, 

and  from  the  relations  of  the  abscissas  and  ordinates  of  points  in 

the  several  quadrants  ;  thus  : 

with  a  given  distance,  to  every  abscissa  correspond  two  ordi- 

nates ; 

therefore  to  every  cosine  correspond  two  sines. 

So,  to  ever}7  ordinate  correspond  two  abscissas  ; 

therefore  to  every  sine  correspond  two  cosines  ;  and  so  on. 

NOTE  3.  Of  the  relations  given  above,  the  most  important  are 
those  between  sine  and  cosecant,  cosine  and  secant,  tangent  and 
cotangent  ; 

those  between  sine  and  cosine,  -tangent  and  secant,  cotangent 
and  cosecant  ; 

and  the  expressions  for  tangent  and  cotangent  as  ratios  of  sine 
and  cosine,  and  for  sine  and  cosine  in  terms  of  tangent. 
These  relations  should  be  committed  to  memory  ;  the  rest  may 
be  proved  as  exercises. 

*  NOTE  4.     The  above  relations  may  also  be  expressed  thus  : 

sin~1a  =  esc"1-, 
a 
_  1  • 

sin"1  (  ±  a)  =  cos"1  V  1  —  a2  =  sec"1    , 


-- 

_  Vl-a2 

wherein  Vl  —  a2  is  either  positive  or  negative  throughout,  inde- 
pendently of  ±a.  The  reader  may  express  in  like  manner  the 
values  of  the  five  other  principal  anti-functions  of  a. 


18 


PLANE   TBIGONOMETBY. 


§  20,     FUNCTIONS  OF   30°,  60°,  120°,  150°,  .    .,  i.e.,  OF        ±     - 

'2         o 

THM.  12.     The  functions  of  —  ±  -  are  : 

2       6 


Angle. 

Sine. 

Cos. 

Tan. 

Cot. 

Sec. 

Csc 

30°,  i.e.,  -;  or2mr  +  - 
6                     6 

i 

W3 

JV3 

V3 

IV3 

2 

'     *'  6  '                *)*  f 

Jv/3 

i 

V3 

W3 

2 

120°,  i.e.,  —  ;  or  (2n  +  2)^  +  ~ 
6                                6 

W3 

-i 

~V3 

-iv3 

-2 

IV 

150°,  i.e.,  —  ;  or(2tt+l)7r-- 
6                                6 

» 

-JV3 

-W3 

-V3 

-IV3 

2 

6  '                             6 

-i 

-iV3 

JV3 

V3 

-iV3 

-2 

240°,le.,  —  ;  or(2»-i)7r-^ 

-ii/8 

2  V° 

_, 

,3 

1.  ,3 

2 

6                                G 

2 

" 

o  V 

300°  fe  107r-  orf2rc     i^4-7r 

t)                                  6 

-JV3 

* 

-V3 

-iV3 

2 

0     •          II71"                                 Ti" 

,t.e.,—  ;o]         -- 

-} 

M 

-W3 

yg 

IV3 

-2 

For,  let  ABC  be  an  equilateral  triangle,                       ^^ 

and  from  A  let  fall  ADJ_BC  ;  then  AD  bisects               s' 

/.  A,  =  60°,  and  side  BC  :                      \qzom.      *,' 

=  30°,     and  DC  =  £  AC. 
But     sin-DAC  =  DC  :  AC  =  J  ; 
.-.   sin  30°=^; 

.-.    cos30°  =Vr=:    = 


The  reader  ma}r  prove  the  remaining  values  as  exercises 
the  preceding  theorems  ;  for  example  : 


upon 


tan  80°  = 


cos  30 


- 

V3 


sin  150°  =  sin  (180°  -  30°)  =  sin  30°  =  £  ; 
sec  330°  =  sec  (-30°)         =  sec  30°  = 


§  21.]          PRIMARY  DEFINITIONS   AND   FORMULAE. 


19 


§  21,     FUNCTIONS  OF  45°,  135°,  i.e.,  OF  mr±  •-. 


THM.  13.     The  functions  of  mr  ±-  are : 

4 


Angle. 

Sine. 

Cos. 

Tan. 

Cot. 

Sec. 

Csc. 

45°,  i.e.,  f;  or  2^  +  7 
4                      4 

'Vi 

.V* 

+  1 

+  1 

V^ 

V2 

135V.e.,—  ;  or  (2?i+l),r-- 
4                              4 

Vi 

-V4 

j 

-1 

-V2 

V2 

225°,  i.e.,  —  ;  or(2»+l)ir+- 
4                              4 

-VI 

-Vi 

+  1 

+  1 

-V-? 

-V2 

315°,  i.e.,—  ;or2tt7r-- 

-VI 

Vi 

-1 

-1 

V2 

-V2 

For,  let  ADC  be  a  right  isosceles  triangle,,  right-angled  at  D  ; 
then  Z  A  =45°, 
and      AD  =  DC  =  AC  .  *J%.  [geom. 

But       sin  A     =  DC  :  AC,  and  cos  A  =  AD  :  AC. 

.  • .     sin  45°  =  A/J,    and  cos  45°  =  ^/-J. 

The  reader  may  prove  the  remaining  val- 
ues as  exercises  upon  the  preceding  the- 
orems ;  for  example : 

tan       45°  =  ^i5!  =  ^t=l; 

«^«  A n°  /I 


sec 


cos  45 

-J— 
cos4o 

°  ° 


tan     225°  =  tan  (180°  +  45°)  =  tan  45°  =  1  ; 
sin     405°  =  sin       45°=      -v/i; 
cos—  405°  =  cos—  45°=      cos  45°  =  ^/^; 
tan     585°  =  tan     225°  =      tan  45°  =  1  ; 
cot  —  585°  =  cot     135°  =  —  tan  45°  =  —  1  ; 
sec     G  75°  =  sec   —45°=      sec  45°  =  V2  5 
esc-  675°  =  esc      45°==   'V2- 


20 


PLANE   TRIGONOMETRY. 


[I- 


§  22,     FUNCTIONS  OF  0°,  90°,  180° ,  i.e.,  OF      TT. 

THM.  14.    The  functions  of  n-rr  are : 


Angle. 

Sine. 

Cos. 

Tan. 

Cot. 

Sec. 

Csc 

0°,  i.e.,  OTT;  or  2mr 

0 

1 

0 

GO 

1 

CO 

90°,  i.e.,  £;   or  (2?i  +  |)7r 

1 

0 

00 

0 

GO 

1 

180°,  i.e.,  TT;    or(2?i-fl)7r 

0 

-1 

0 

oo 

—  1 

00 

270°,  i.e.,  ^;  or(2n-i)B- 

-1 

0 

GO 

0 

GO 

-1 

The  reader  may  prove  these  values  by  direct  reference  to  the 
definitions  of  the  functions.  [§  7 

NOTE.  From  the  values  given  above,  and  from  others  com- 
puted by  methods  to  be  given  later,  tables  have  been  constructed, 
giving  the  functions  of  all  angles,  and  other  tables  giving  the 
logarithms  of  these  functions ;  the}7  are  called  trigonometric 
tables.  For  their  use  the  reader  is  referred  to  the  tables  them- 
selves. 

§  23,     GRAPHICAL  REPRESENTATION  OF  FUNCTIONS. 
Line-Functions. 


Let  XOP  be  any  angle,  and  with  o  as  center,  and  any  radius 
ox,  describe  a  circle,  cutting  the  sides  in  x  and  P  ;  from  p  let 


§  23.]          PRIMARY  DEFINITIONS   AND   FORMULAE.  21 

fall  PA_Lox  ;  at  x  draw  XT  tangent  to  XP,  and  meeting  OP  in  T. 
Draw  OY_Lox,   and  meeting  the   circle  at  Y;    from  Y  let  fall 
YB_Lop,  and  at  p  draw  PT'  tangent  to  PY,  and  meeting  OY  in  T'  ; 
then  :  if  the  radius  be  taken  as  the  unit  of  length,  the  ratio  AP  :  ox 
is  the  numerical  measure  of  AP  ;  i.e.,  the  number  of  units  in  the 
length  of  the  line  AP  is  equal  to  sinxop, 
and        the  line  AP  represents  sin  XOP. 
So,         the  line  OA  represents  cos  XOP  ; 

the  line  XT  represents  tan  XOP  ; 

the  line  OT  represents  sec  XOP  ; 

the  line  AX  represents  versxor. 
So,         the  line  BY  represents  sin  POY,  i.e.,  cos  XOP  ; 

the  line  PT' represents  tan  POY,  i.e.,  cot  XOP  ; 

the  line  OT' represents  secpOY,  i.e.,  CSCXOP  ; 

the  line  BP  represents  versPOY,  i.e.,  covers  XOP. 

These  lines  ma}'  be  called  the  line-functions  of  the  angles,  as 
distinguished  from  the  ratio-functions  heretofore  defined.  They 
are  also  called  functions  of  arcs;  and  the  others,  functions  of 
angles. 

In  most  of  the  earlier  treatises  on  trigonometry  the  functions 
were  always  defined  as  lines,  thus  :  the  sine  was  said  to  be  "the 
perpendicular  from  one  extremit}'  of  an  arc  upon  the  diameter 
through  the  other  extremity";  the  cosine  was  "the  distance  from 
the  center  to  the  foot  of  the  sine,"  or  "  the  sine  of  the  comple- 
ment," and  so  on. 

NOTE  1.  These  line-functions  must  never  be  thought  of  as 
identical  with  the  ratio-functions,  not  even  when  the  radius  is 
unit}* ;  the  ratio-functions  are  neither  lines  nor  the  lengths  of 
lines,  but  numbers  merely  whose  values  are  independent  of  any 
measuring  unit  of  length.  The  reader  may  name  the  line-func- 
tions in  the  different  figures,  may  state  which  of  them  are  positive 
and  which  negative,  and  may  show  that  the  ratios  of  these  line- 
functions  to  the  radius  are  identical  with  the  ratio-functions 
heretofore  given. 


22 


PLANE  TRIGONOMETRY. 
Curve  of  Sines. 


[I- 


Let  ox  be  the  radius  of  a  circle,  and  divide  the  circumference 

into  any  convenient  parts  at  PI?  P2,  ;    draw  A^,  A2p2,  , 

ordinates  of  P15  P2, with  reference  to  ox,  and  sines  of  the  arcs 

XP15  xp2,  .....;  draw  XY  tangent  to  the  circle  at  x. 

Conceive  the  circle  to  roll  along  XT  ;  let  B15  B2, be  the  points 

on  XY  where  PJ,  P2,  rest  respectively;  and  at  BI?  B2,  erect 

perpendiculars  to  XY,  and  make  B^  =  A^,  B2c2  =  A2p2, 

Through  c1?  C2, draw  a  smooth  curve  ;  it  is  the  curve  of  sines. 

NOTE  2.     The  following  relations  are  manifest : 

(1)  At  x  the  sine  is  0. 

(2)  While  the  arc  is  small,  the  sine  is  nearly  as  long  as  the  arc. 

(3)  The  sine  increases  more  and  more  slowly. 

(4)  It  is  equal  to  + 1,  its  maximum,  when  the  arc  is  a  quad- 
rant ;  then 

(5)  It  decreases,  at  first  very  slowly,  but  faster  and  faster  as 
the  arc  approaches  a  half-circle. 

(6)  When  the  arc  is  a  half-circle,  the  sine  is  0. 

(7)  While  the  arc  increases  from  180°  to  270°,  the  sine  de- 
creases from  0  to  —1,  its  minimum. 

•(8)  While  the  arc  increases  from  270°  to  360°,  the  sine  in- 
creases from  —1  to  0. 

(9)  At  the  end  of  the  first  revolution  the  sine  is  again  0. 

(10)  During  every  successive  revolution  the  same  phenomena 
are  repeated  in  the  same  order,  and  for  negative  as  well  as  posi- 
tive arcs. 

(11)  While  the  revolution  is  continuous,  the  values  of  the  sines 
are  periodic,  every  successive  revolution  indicating  a  new  cycle, 
and  a  new  wave  in  the  curve. 


§24.]  1MIIMAUY    DEFINITIONS   AND   FORMULAE.  23 

(12)  The  four  parts  of  each  wave  that  correspond  to  the  four 
quadrants  of  the  angle  XOP,  are  equal  and  similar  to  each  other. 

(13)  The  sine  has  no  value  greater  than  -f- 1  •>  nor  less  than  —  1 . 
The  reader  ma}'  draw  curves  to  represent  the  other  functions, 

and  discuss  them  ;  lie  will  find,  among  other  things,  that : 

(14)  The  tangent  is  0  at  0  =  0  ; 
increases  through  the  first  quadrant  to  +  °°  I 
at  90°  changes  suddenly  to  —  oo  ; 

increases  through  the  second  quadrant  to  0  at  180°  ; 

increases  through  the  third  quadrant  to  +  oc  at  270°  ; 

at  270°  changes  to  —  oo  ; 

increases  through  the  fourth  quadrant  to  0  at  360°  ;  and  so  on. 

(15)  The  secant  is  +1  at  0  =  0  ; 
increases  through  the  first  quadrant  to  +  °°  ; 
at  90°  changes  to  —  oo  ; 

increases  through  the  second  quadrant  to  —1  at  180°  ; 

decreases  through  the  third  quadrant  to  —  oo  at  270° ; 

at  270°  changes  to  -f-  oo ; 

decreases  through  the  fourth  quadrant  to  +1  at  360°  ;  and  so  on. 

It  has  no  value  between  + 1  and  —  1 . 

(16)  The  cosine,  cotangent  and  cosecant  have  the  same  limits 
as  the  sine,  tangent  and  secant  respectively ;  they  go  through 
like  changes  and  are  represented  b}T  like  curves  ;  but  they  begin, 
for  0  =  0,  with  different  values,  viz.,  1,  oo,  and  oo. 

§  24,     EXERCISES. 

1 .  Express  in  degree-measure  the  angles  : 

7T       7T       5  73 

5'    7'    9'    ~3~ 

2.  Express  in  7r-measure  the  angles  : 

14°,  15°,  24°,  120°,  137°  15',  -4800°,  13',  24",  -5°,  19'  37".o. 

3.  If  the  radius  of  a  circle  be  one  inch,  what  is  the  length  of 
the  arcs : 

14°,    15°,    120°,    57°17'44".8,    1°,    1',    1",    ?,   £,   2,  *r+l. 

5     7 


PLANE   TRIGONOMETRY.  [I. 

4.  Find  the  complements  and  the  supplements  of  the  angles  : 

37°,   215°,   325°,   107°  12' 15",   -36°  12',   £     ^     !!5     _^5 

o'     7  '      9   '         3  ' 

5.  Construct  the  points  in  a  plane  whose  abscissas  and  ordi- 
nates  are : 

°i  3  I     2,9;     8,   —7;     —3,4;     —5,   —8.      [Use  any  convenient  unit.] 

6.  Construct  the  points  in  a  plane  whose  bearings  and  dis- 
tances are : 

30°,  10;  30°, -10;  210°,  10;  -150°,  10;  150°, -10;  -GO0, 10; 

TT,  8  ;  TT,  —8  ;   — TT,  8  ;  0°,  8  ;  — ,  4  ;  — ,  -4  •  — -  4  •  — ?T    _/L 

3'         3'  3'  3'" 

7.  Construct  the  angles  whose 

sines  are  -J,    -f,    V^pl ?    -^    Oj    i  +  V3t 

cosines  are       .6,    ±-£,    ±j-ys,    Lz_V?. 

2 
tangents  are      £,    f,    0,    -1,    oo ,    V2+1. 

cotangents  are  i,   3^,   2^  ~  6,  [o  and  6,  any  lines. 
0  -yo 

•*  8.    Write  formulae  for  all  values  of  0  : 

when  sin  0  =  —  sin  a,  Vsin2a,  VcosV. 
when  cos 6)  = —cos a,  Vcos2a,  Vsin2a. 
when  tan 9  =  —tana,  Vtan2a,  Vcot2a. 
when  cot  0  =  —  cot  a,  Vcot2a,  Vtan2a. 
9.  Find  the  remaining  functions  of  6  : 

if  sin  0=. 6,   — .   ?,    -?,    J_     J^-L  V3 
17     4         3      -v/5'    V2        6 


,  , 

2  '    13'      2V2 


yil        z  a  a 

"3~"'    1-a2'    ~"b' 


ifcsc^=-2,   |,   t±^2,   ^,    2(-l)".  [m  any  integer. 


§  24.]          PRIMARY  DEFINITIONS   AND   FORMULAE.  25 

*]0.    Find  sin0  from  the  equation,  sin 0. cos 0  =  m. 

*11.    If  tan0  +  cot0  =  ra,   express  all  the  functions  of  0  in 
terms  of  m. 

*12.    If  tan  0=  f-Y,  find  the  value  of  —  +  _A_. 
\aj  cos0      sin0 

*13.  Eliminate  0  from  the  equations  : 

sin0-f-cos0  =  a,  tan0  +  sec0  =  2&. 

*  14.  Eliminate  <£  from  the  equations  : 

esc  (f>  —  sin  <£  =  m,  sec  <f>  —  cos  </>  =  n. 

*15.    Eliminate  0  and  $  from  the  equations : 


asin20-f&cos20  =  m,    6sin2</>4-acos2<^  =  ?i,    a  tan  0  =  b  tan  <£. 
*16.    If  tan<^-hsec<^  =  a,  find  si 


*17.    Show  geometrically  that  sin 2  <£ < 2 sin <£,  if  <£>0,  <TT. 

*18.    Show  how  to  divide  an  angle  into  two  parts  which  shall 
have  their  sines  in  a  given  ratio ;  their  cosines  in  a  given  ratio. 

*19.    Show  how  to  construct  an  angle  whose  tangent  is  four 
times  its  sine. 

*20.   Find  all  the  angles  whose 

sines  are         ±Vi'    i^V^  sin 50°,  cos 50°,  cos—. 

16 

cosines  are     ±Vi>    ^iV^*  cos50°,  sin  50°,  sin~. 

16 

tangents  are  ±1,       ±V^>     tan-,     cot  40°,  cot  ^. 

8  8 

cotangents  are  0,    '  ±^3,     cot  —  ,  tan  20°,  tan-^- 

8  8 

secants  are        oo  ,     ± — ,     sec  ^x  esc  35°,   esc -^- 

\fo  8  8 

7  7T  7  7T 

cosecants  are    oo  ,     ±2,        esc  •— ,  sec  35°,   sec  — - • 

8  o 


26  PLANE   TRIGONOMETRY.  [I. 

21.   Find  the 

•  97-w 

sine  of  225°,     —585°,     810°,     —960°,     STT,     —  —  . 

4 

cosine  of        315°,     -675°,     960°,   -1110°,    ^,    -877. 


tangent  of     495°,     -945°,   1110°,   -1260°,     —  ,  -515. 

6  4 

cotangent  of  675°,  -1035°,   1260°,  -1410°,     —,  -l^7". 

G  4 

secant  of       855°,  -1215°,   1410°,   -1560°,     —  ,  -41?. 

6  4 

cosecant  of  1035°,  -1395°,   1560°,  -1710°,     ^5,  -13  TT. 

6 

22.    In  terms  of  the  functions  of  positive  angles  less  than  90°, 
express  the  values  of  the 

sine  of  135°,     335°,   -535°,     -735°, 


<0  QQKO  -orO  r-QKO  27?T  29  7T 


5  ' 

cosine  of        235°,     435°,   -635°,     -835°,      ?^5, 

O 

tangent  of      335°,      535°,   -735°,     -935°,      —,    -  — 

0  7 

QO  _ 

cotangent  of  435°,      635°,   -835°,  -1035°,         ^,    -OTT. 

o 

secant  of        535°,      735°,   -935°,  -1135°, 

cosecant  of    635°,      835°,  — 1035°,  -1235°,         ,    , 

5  7 

23.    In  terms  of  the  functions  of  positive  angles  less  than  45° 
express  the  values  of  the 
sine  of  50°,      150°,    -250°,    -350°,     ?f,    -4ir. 


°  ° °  ° 


cosine  of         60°,      160°,    -260°,    -360°,       5,  -^ 

12  o 

tangent  of       70°,      170°,    -270°,    -370°,     ^T,  - 

12  o 

cotangent  of    80°,      180°,    -280°,    -380°,          ,  -677. 


§  24.]          P1UMAHY   DEFINITIONS   AND   FORMULAE.  27 


secant  of   90°,  190°,  -290°,  —390°,  -~,  -    . 

cosecant  of  100°,  200°,  -300°,  -400°,  —  ,  -  —  . 

12  3 

*  24.    Trace  the  changes,  when  6  increases  from  0  to  2?r,  in  tho 
sign  and  value  of  each  of  the  expressions  : 
sin  0  +  cos  0)        sin  6  —  cos  0,        sin  0  +  esc  0,        tan  0  —  cot  0, 
sin20,     cos20,     sin20  —  cos20,     cos2$  —  sin2(9,     tan20-f  cot20. 

25.  From  the  table  of  natural  functions,  find  the 

sin  of  20°,  21°,  20°  10',  20°10'45",  89°18'25",  157°15'23". 

cos  of  20°,  21°,  20°  10',  20°10'45",  89°18'25",  157°15'23". 

tan  of  35°,  36°,  35°  15',  35°15'47",  89°58'35",  125°   0'12". 

cot  of  35°,  36°,  35°  15',  35°15'47",  89°58'35",  125°   0'12". 

26.  From  the  table  of  natural  functions  find  the  angles  whose 
sines  are  .25882,  .25910,  .25900,  .92794,  .92805,  .92800. 
cosines  are        .92794,  .92805,  .92800,  .25910,  .25882,  .25900. 
tangents  are      .5022,    .5059,    .5035,     .9217,     .9271,     .9250. 
cotangents  are  .9217,    .9271,     .9250,    .5022,    .5059,     .5035. 

27.  From  the  table  of  logarithmic  functions  find  the  logarithmic 
sin  of  20°,  21°,  20°  10',  20°10'45",  '   89°18'25",  157°15'23". 
cos  of  20°,  21°,  20°  10',  20°10'45",      89°18'25",  157°15'23". 
tan  of  35°,  36°,  35°15',  35°15'47",      89°58'35",  125°   0'12". 
cot  of  35°,  36°,  35°  15',  35°15'47",      89°58'35",  125°   0'12". 
sec  of  50°,  51°,  50°20',  50°20'49",    115°   0'45",  179°58'55". 
esc  of  50°,  51°,  50°  20',  50°20'49",    115°-0'45",  179°58'55". 

28.  From  the  tabled  of  logarithmic  functions,  find  the  angles 
whose  logarithmic 

sines  are             8.580892,  8.584193,  8.582125,  9.999683. 

cosines  are         8.580892,  8.584193,  8.582125,  9.999683. 

tangents  are       8.581208,  8.584514,  8.583125,  11.418790. 

cotangents  are  8.581208,  8.584514,  8.583125,  11.418790. 

secants  are       10.367529,  11.367514,  12.367529,  13.367514. 

cosecants  are   10.367529,  11.367514,  12.367529,  13.367514. 


28 


PLANE   TBIGOXOMETBY. 


[II. 


II.     GENERAL   FORMULAE. 

§  1,    FUNCTIONS  OF  THE  SUM,  AND  OF  THE  DIFFERENCE,  OF 
TWO  ANGLES. 


THEOREM  1.    If  Q  and  0'  be  any  two  angles,  then  : 


39] 
40] 
41] 
42] 


sin  (0  +  0')  =  sin  0  cos0'+  cos  0  sin  0'  ; 
sin  (0  —  6')  =  sin  0  cos0'—  cos  0  sin  0'  ; 
cos(0+  0'}  =  COS0COS0'-  sin<9  sin0'  ;     / 
cos(0-  0')  =  cos  0  cos  0'+  sin  0  sin  0'. 


D    ?/__.         _K 


Fx'Q 


For,  let  XOP  be  any  angle  0,  positive  or  negative,  and  POQ  be 
an}'  other  angle  0'  ;  then  XOQ  =  0  +  0'. 

Draw  BQ  ordinate  of  Q.  with  reference  to  OP,  and  AQ  and  CB  ordi- 
nates  of  Q  and  B  with  reference  to  ox  ;  draw  BK  parallel  to  ox  and 
meeting  AQ  in  D  ;  on  BQ  take  F  so  that  Z  KBF  =  £TT  -f  0.  Then  : 

=  ordinate  AQ  =  CB  +  DQ  .  [L§7 

distance  OQ          OQ 


(1) 
but 


and 


OQ        OB     OQ 

=  sin#cos0f, 
£Q  =  £S.5Q 

OQ        BQ     OQ 


=  cos0sin0'; 


[I.  §7 

[I.  §7 

[7 

Q.  E.  D. 


§  1.]  GENERAL  FORMULAE.  29 

(2)  In  [39]  substitute  —0'  for  0',  then: 
...     0  -6'  =0  +  (-&), 

.-.     sin  (0  —  6')  =  sin0cos(-0')  +  cos0sin(-0')  [39 

=  sin  0  cos  0'—  cos  6  sin  0'.         Q.E.D.    [1,2 


(3)        cos(0  +  0Q  =  =  ;  [i.  §7 

distance  OQ          OQ 
but  2°  =  °^.°^ 

OQ        OB     OQ 

=  cos0cos0',  [I.  §7 

and  BD  =  BD.BQ 

OQ        BQ     OQ 

.   =  cos(i7r  +  0)sin0'  [I.  §7 

=  —  sin0sin0';  [8 

=  cos0cos0'—  sin0sin0'.  Q.E.D. 


(4)  In  [41]  substitute  — 0'  for  0',  then 

cos  (0  -  0')  =  cos  0  cos  (—0')  —  sin0  sin(—  0')  [41 

=  cos0cos0'+sin0sin0'.        Q.E.D.     [2,1 

NOTE.  Since  each  of  the  angles  0  and  0'  may  be  in  either  of 
the  four  quadrants,  there  are  sixteen  different  cases  possible  ;  but 
the  proof  is  general,  for  it  applies  to  all  alike.  The  reader  may 
enumerate  the  cases  in  detail  and  draw  the  other  twelve  figures. 

COR.  1.     IfO  and  0'  be  any  two  angles,  then: 
43]  tan  (0  -}-  0')  =  - 

/A      /if\       tan  0  — tan  0' 

44  I  tan  (0  —  0')  =  —  -. 

l  +  tan0tan0' 

[35 


_  sin  0  cos  0'+  cos  0  sin  0'  p™    .  . 

~cos0cos0'~  sin  0  sin  0'* 
Divide  both  terms  of  the  fraction  by  cos  0  cos  0',  then 


^  _  [35 

1  —  tan  0  tan  0' 


30  PLANE   TRIGONOMETRY.  [II. 

So'       ""^-SgEg  P» 

_  sin  0  cos  0'  —  cos  0  sin  0f 
cos  6*  cos  0'  H-  sin  0  sin  0' 
tan  0-  tan  0' 
~l  +  tan0tan0'' 
The  reader  may  find  like  formulae  for  cot  (0+6')  and  cot  (0—0')  • 

NOTE.  The  six  formulae  [39-44]  may  be  written  as  three  : 

39,  40]  sin  (0  ±  0')  =  sin  0  cos  0'  ±  cos  6  sin  6'  ; 

41,  42]  cos  (0  ±  0')  =  cos  0  cos  0'  q:  sin  0  sin  0'  ; 

43,44]  tan(0±0')  = 


COR.  2.     7jf0  and  0'  6e  a?iy  ^wo  angles,  then: 
45] 

46] 
47] 


cos  6»  cos  0' 
48]  £251^1 

COS  0  COS  0' 

49] 


50]  .=cot^          gr 

sin  0  cos  0; 

The  reader  may  prove  these  formulae  by  performing  the  di 
visions  and  reductions  indicated. 

COR.  3.     I/O  and  O1  be  any  two  angles,  then: 
51]  sin  (0  +  O1)  +  sin  (0  -0')=     2  sin  0  cos  6'  ; 

52]  sin  (0  +  0')  -  sin  (0  -  6')  =     2  cos  6  sin  6'  ; 

53]  cos  (0  +  0')  +  cos  (0  -  0r)  =     2  cos  0  cos  0'  ; 

54]  cos  (0  +  0')  -  cos  (0  -  0')  =  -  2  sin  0  sin  0'. 


§  2.]  GENERAL  FORMULAE.  31 

For,  if  to  [39],  [40]  be  added,  the  result  is  [51]  ; 
if  from  [39],  [40]  be  subtracted,  the  result  is  [52]  ; 
if  to  [41],  [42]  be  added,  the  result  is  [53]  ; 
if  from  [41],  [42]  be  subtracted,  the  result  is  [54]. 

COR.  4.    I/O  and  0'  be  any  two  angles,  then: 
55]        sin  (0  +  .0')  sin  (0  —  6')  =  sin20  -  sin20'=  cos2<9'—  cos20  ; 
56]        cos(0  -f-  6')  cos(0  -  6')  =  cos20-  sin20'=  cos20'—  sin2<9. 

The  reader  may  prove,  by  performing  the  multiplications  indi- 
cated ;  he  will  make  use  of  [36]  . 


COR.  5.    If  6  and  0'  be  any  two  angles,  then: 


sin  (0  +#')  _  tan  0  +  tan  0f . 
sin  (0-0')  ~~  tan  0  -  tan  0' ' 
cos  (0  +0')  _  1  -  tan  0  tan  0' 
cos  (0  -  0')  ~  1  -f  tan  0  tan  0' 
sin  (0  ±  0')  tan  0  ±  tan  0' 


57] 


591 

™/7».i-/i»\      I±tan0tan0' 


The  reader  may  prove,  by  aid  of  [47-48]. 

§  2.     FUNCTIONS  OF   DOUBLE  ANGLES,   AND   OF   HALF 
ANGLES. 

THM.  2.    If  0  be  any  angle,  then: 
60]        sm20  =  2sin0cos0; 
61]        cos  2  0  =  cos20  —  sin20  =  2  cos20  —1  =  1  —  2sin20  ; 


not  -1/1                  fl    —  COS0 

63]  sm|0  =  ^|  -  -  -- 

„.-<  -,  /,             /1  +  COS0 

64] 


—  cos0         sin0          1  —  C030 
65] 


32  PLANE   TRIGONOMETRY.  [II. 

(1)  In  [39,  41,  43]  substitute  0  for  0',  then: 

•.-      0+0'  =2(9; 

.-.      Sm20  =  sin0cos0-f-cos0sin0  =  2sin0cos0.  [39 

So,          cos20  =  cos0cos0  —  sin0sin0='cos20  —  sin20  [41 

=  2cos26>  -1  =  1  -  2sin20  ;  [36 


and         tan20  =  -  tan0          2tan0  E>  D> 

1—  tau0tan0      1—  tan2(9 

(2)-.-      cos20  =  l-2sin20,    whatever  the  angle  (9,  [61 

.-.      cos0    =  !-2sin2iJ-0,  [substitute  £0  for  0 

[1-COS0 

•'•      sm-J0  =  ^j  --  £- 

So,  •  .  •      cos  2  0  =  2  cos  2  0  —  1  ,    whatever  the  angle  0,  [61 

.-.      cos0    =2cos2£0—  1,  [substitute  |0  for  0 

1+COS0 


Hence,    tani0  =  =  A  l1-^^.  [35 

cos^-0      \l+cos0 


Multiply  both  terms  of  the  fraction  by  ^/(l  -f  cos0)  ,  then 
tan£0  =,  Va-eo8'g)  =     sin0 

1+COS0  1+COS0 

or,  multiply  both  terms  of  the  fraction  by  ^/(l  —  cos0)  ,  then 


NOTE.    Although  the  radical  ^/(l  —  cos20)  ,  =  ±sin0,  has  two 
values,  opposites  ;  yet  tan  -J-0  has  but  one  sign  and  one  value  ; 
for,  when  0  is  in  the  first  or  second  quadrant,  then  4-0  is  in  the 
first  or  third  quadrant,  and  sin0  and  tan  J-0  are  both  positive  ; 
so,  when  0  is  in  the  third  or  fourth  quadrant,  then  |-0  is  in  the  sec- 
ond or  fourth  quadrant,  and  sin0  and  tan  J0  are  both  negative  ; 
i.e.  ,  whatever  the  sign  of  sin  0,  the  same  is  that  of  tan  £  0.  [I.  Th.  3 

COR.    IfObe  any  angle,  then: 
66]          sin£0  =  ![V(l+sin0)-V(l-sin0)]; 
67]          cos  J0  =  J  [V(l  +  sin0)  +  %/(l-  sin0)].  [See  nt.  1  p.  103 


§3.J  GENERAL  FORMULAE.  33 

For,  • .  •  sin  2  0  .          =  2  sin  0  cos  0,  whatever  the  angle  0,      [60 

.  • .  sin  0  =  2  sin  1 0  cos  -j-  0  ;  [substitute  £  0  for  0 

but  1  =  cos2 £04  sin2 £0;  [36 

.-.  l+sin0        =  (cos  ^0  +  sin  |0) 2, 

and  1— sin0        =  (cosJ0  —  sinJ0)2; 

.-.  V(1  +  sin0)  =  cos  |0  +  sin£0, 

and  VC1  —  sin0)  =  cos  i#  — sin£0. 

.•.  etc.  Q.  E.  D 

§  3,    FUNCTIONS  OF  THE  HALF-SUM,  AND  OF  THE  HALF- 
DIFFERENCE,  OF  TWO  ANGLES. 

THM.  3.    IfO  and  0'  be  any  two  angles,  then: 
68]  sin  0  + sin  0'=      2 sin  |(0  +0')cosJ(0  —  0')  ; 

69]  sin0  —  sin0'=      2cosi(0  +  0')smJ(0  — 0')  ; 

70]  cos 0  + cos 0'=      2cos^(0  +  0')cosi-(0-0')  ; 

71]  cos0  -  cos0'=  -  2 sin  |(0  +  0')sin  J(0  -  0r) . 

For,  in  [51-54]  substitute  |(0  +  0')  for  0  and  i(0  —  0')  for  0f ; 
then  also  0  stands  for  (0  +  0'),  and  0'  for  (0  -  0')  ; 
and  [51-54]  reduce  to  [68-71]  respectively. 

COR.     If  6  and  0'  be  any  two  angles: 


f"J 
73] 

74] 
75] 
76] 

771 

sin  0  —  sin  0' 
sin  0  4-  sin  0' 

tan  £  (0  4-  0')  ; 
tan  i  (0-0'); 

pot  1  f/J  d'"\  • 

cos  0  +  cos  0' 
sin  0  —  sin  0' 

cos0  4-  cos0' 
sin  0  4-  sin  0f 

cos  0  —  cos  0' 
sin  0  —  sin  0' 

COl  -j  ^(7  "  /  J 

cot  i  (040'); 

cos  0  —  cos  0' 
cos0  4-  cos0r 

The  reader  ma}r  prove,  by  dividing  [68-71]  one  by  another, 
as  indicated,  and  reducing  the  quotients. 


34  PLANE   TRIGONOMETRY.  [II. 

*§  4.   FUNCTIONS  OF  THE  SUM  OF  THREE  OR  MORE  ANGLES, 
OF  TRIPLE  ANGLES,  ETC. 

THM.  4.    If  0,  0',  0",  .....  be  any  angles,  then: 

781       sin  (04-0'+  0"^=J      sill<9  cos^'cos^'  +  cos  0  sin  0'  cos  0" 
1  +  cos  0  cos  0'  sin  0"  -  sin  0  sin  0'  sin  6" 


79]       cos(0+0'+  0")  =  (     c08*008*'009*"  -  eos  6  sin  0'  sin  0" 
}      l-sin0cos0'sin0"-sin0sin0'cos0" 

=  cos0cos0'cos0"(l  —  tan  0'  tan  0"—  tan0tan0"—  tan  0  tan  0'); 

801       tanffl  I  0f  I  g^=tang  +  tangf  +  tangff-t»ngtangftangfy 
l-tan0'tan0"  —  tan  0  tan  0"-  tan  0  tan  0'' 

The  reader  may  prove,  by  developing  sin  [0  +($'  +  0")]? 
cos[0-h(0'+0")],  and  tan[0+(0'+0")]  as  functions  of  the  sum 
of  two  angles  0  and  (0'+0"),  and  then  developing  the  functions 
of  (0'+0")  which  are  involved.  He  may  also  prove  the  result  for 
tan  (0  4-  0'  +  0")  by  dividing  that  for  sin  (0  -f  0'  +  0")  by  that 
forcos(0  +  0'  +  0"). 

The  reader  ma}*  in  like  manner  get  the  sine,  cosine,  and  tan- 
gent of  the  sum  of  four  angles,  of  five  angles,  and  so  on.  For 
sine  and  cosine,  he  will  find  that  : 

2tan0-2tan0tan0ftan0" 


821        cos(0+0f+0ff+  .....  )_f      1  -  Stan  0  tan  0' 

cos  0  cos  0'  cos  0"  .....       !-j-2tan0tan0'tan0"tan0'"-  .....  ; 

wherein  2tan0  stands  for  the  sum  of  the  tangents  of  0,  0',  .....  , 
2  tan  0  tan  0'  stands  for  the  sum  of  the  products  of  those  tangents 
taken  two  and  two  ;  and  so  on. 

By  making  0  =  0'  =  0"  =  .....  ,  he  will  find  that  : 
83]        sin  3  0  =  3  sin0  cos20  -  sin30 

=  3sin0-4sin30;  [36 

84]        cos30  =  cos30-3cos0sin20 

=  -3cos0-f  4cos30; 


§  5.]  GENERAL  FORMULAE.  35 

85]      sin  40  =  4sin0  cos30  —  4sin30  cos0 

=  4sin0cos0  — 8sin30cos0;  [36 

86]      cos40  =  cos40  —  Gcos20sin2<9-f-sin40 

=  1  —  8cos20  +  8cos40; 
87]      sin  50  =  5 sin<9 cos40  -  10sin3<9  cos20  -f  sin5(9 

=  5sin0  -  20  sin3  0  +  lGsin50  ; 
88]      cos50*=cos50-10cos30sin20  +  5cos0sin40 

=  5cos0  -  20cos30  +  16cos50  ; 
and.  in  general,  that : 
89]      sin  n  0  =  cn>1  sin  0  cos""1^  —  cn)3sin30  cosn~3  0 

+  cn)5sin50cosn-50  — ; 

90]      cosn0  =  cosn0-cn,2sm20cosn-20+cn)4sin40cosr:-40-- ; 

wherein  cn>r  denotes 

n(n  —  1)  (71  —  2) (n  —  r  +  1) 

1.2.3 r 

the  number  of  combinations  of  n  things  taken  r  at  a  time. 

*§5,    DIFFERENTIATION  OF  TRIGONOMETRIC  FUNCTIONS. 

LEMMA.    If  6  be  an  infinitesimal  angle,  then  lim  (sin0  :  0)  =1, 
and  lim  (tan  0  :  0)  =  1 . 

For,  let  c  be  the  circumference  of  a  circle, 
o  the  centre,  p  and  j>r  the  perimeters  of  two 
regular  polygons  of  the  same  number  of  sides, 
the  first  inscribed  in  and  the  other  circum- 
scribed about  the  circle,  and  having  their 

sides  PP',  TT', parallel  each  to  each. 

Draw  OAX_LPP'  and  TT',  then  : 

*.*  2)<c<P'->  whatever  the  number  of  sides, 
and  •.•  p  approaches  indefinite^-  near  to  p1  when  the  number  of 
sides  is  indefinitely  increased,  [geom. 

.*.  c  is  the  common  limit  of  p  and  p', 
.•.   1  is  the  common  limit  of  the  ratios  p :  c  and  p' :  c. 


36  PLANE   TRIGONOMETRY.  [II. 

And-.-      the  two  polygons  are  similar,  and  AP,  XP  and  XT  are 
like  parts  of  p,  c  and  p\  [geom. 

AP  :  XP     =  p  :  c,    and  XT  :  XP  =  p1  :  c  ; 


Vrp         "X^T* 

and  tan  6:0,  =  —  :  —  =  XT  :  XP,  =  p'  :  c  ; 

ox   ox 

.-.      Iim(sin0  :  0)  =  lim  (p  :  c)  =  1, 
and  lim  (tan  0  :  0)  =  lim  (p1  :  c)  =  1  . 

THM.  5.     If  0  be  any  angle,  then: 
91]  D0Sin0  =      cos0;         D0csc0  =  —  cot0csc0  ; 

92]  D0Cos0  =  —  sin  0;         D0sec0=      tan0sec0; 

93]  D0tan^=      sec2^;       T>ecotO  =  —  csc20. 

For,  let  0  be  any  angle,  and  0'  an  infinitesimal  angle,  the  incre- 
ment of  0,  then  : 

v      sin  (0  +  6')  -  sin0  =  2  cos  (0  +  -J0')  sin  Jtfr  ;  [69 


0'  Y</ 

But  0'  is  the  increment  of  0,  [hypoth. 

and  sin  (0+0')— sin0  is  the  corresponding  increment  of  sin  0. 

limincsm  ,  =  Dflsin0,  =  cos0.  [lemma 

inc0 

So,    •••      c< 

t-f-          -=-slaV^^;—  -, 


..  ) 

inc0 

sin  0      cos  0  -DO  sin  0  —  sin  0  D0  cos  0 
So,  an0  =  D9—    =  -  -- 


COS20 

The  reader  may  prove  for  D0  esc  0,  =  D0  —  —  -  ,   for  ve  sec  0 

sin  0 
and  for  DnCot0. 


§  6.]  GENERAL  FORMULAE.  87 

*§  6,    DEVELOPMENT  OF  TRIGONOMETRIC   FUNCTIONS. 
THM.  6.     If  0  be  any  angle,  then: 
94] 


95] 

For  assume 


wherein  A,  B,  c,  .....  are  unknown,  but  constant,  whatever  the 

value  of  6  ;  and  differentiate  both  members  of  the  equation,  then 
cos0  =  B-}-2c<9  +  3D<92+4E03+5F04+  .....  .          [91 

So,         —  sin0=2c  +  GD0-|-12E02-f  20r03+  .....  ;  [92 

—  cos0  =  GD  +  24E0-f  GOF02-f  .....  ; 
sin  0  —  24E  +  120F0  -f  .....  ; 

COS0=120F'-f  .....  . 

Let  0  =  0,  then  : 

sin0  =  A;  .-.  A  =  0  ; 

COS^  =  B  ;  .-.  B  =  1  ; 

—  sin0=  2c  ;  .-.  c  =  0  ; 

3!o;  .-.  D  =  —  —  ;• 

4!E;  .-.  E  =0; 

5!F|  ...  F  =  J_; 

^  5! 

and  so  on. 

Substitute  their  values  for  A,  B,  c,  .....  in  the  assumed  develop- 
ment, then 


--^  .....  . 

Differentiate  both  members  of  this  equation,  then 

/)2          /J4          /J6 

,-+L-L  +  .....  . 


NOTE.  That  sin0  =  A  +  B0  +  c02+D03+  .....  ,  as  assumed  in 
the  above  demonstration,  is  not  self-evident,  though  true.  Most 
functions  ca-n  be  developed  in  this  form,  but  some  cannot. 


38  PLANE   TRIGONOMETRY.  [II. 

COR.     If  6  be  any  angle,  then: 

)5  17/37 


»UJ 

97] 
98] 
99"! 

lilllV 

nrtt  ft  — 

U-f 

1 
0 

1  + 

1  , 

3    ' 
0 
3 

3.5 

03 

1  32. 

2 

5.7 

-f-  — 

• 

5  . 

--h  ; 

/ 

sec0  = 
cse0  — 

32. 

5 

33.5.7 
6106 

33 
2 

310 

770 
.32. 

5 

7 

8 
_+.   ,..; 

"2,  °  • 
4- 

8 
7 

24. 

32.5 

4-  • 

The  reader  may  prove,  as  exercises,  b}^  either  of  two  methods  : 
(  1  )  By  Division  : 

For  the  tangent,  divide  the  development  of  the  sine  by  that  of 
the  cosine  ; 

for  the  cotangent,  divide  the  development  of  the  cosine  by  that 
of  the  sine  ; 

for  the  secant,  divide  unity  by  the  development  of  the  cosine  ; 
for  the  cosecant,  divide  unity  by  the  development  of  the  sine. 
The  work  is  facilitated  by  using  detached  coefficients. 

(2)  By  the  method  of  unknown  coefficients: 


Assume  tan<9,  = 


1—    ^J.6'  2 

+~" 


and  thence  find  A,  B,  c,  ..... 

So,  for  cotangent,  secant,  and  cosecant. 


D0  °»S1  gin(9  0      3      32.5 

/\  •?  S\  A.  s\R 

100]      .-.  Iogsin0      = 


2.3      22.32.5      34.5.7 
i.e.,  the  expression  whose  T>Q  is  the  above  series  for  cot0. 

ciri  /3  /33    ' 

So,        '.'-DQlogcosO^--  -  =  -tan0  =  -(0  +  —  +  .....  ),[96 
cos0  3 


101]     ,.logcoSe     .==-g  +  |l3+-|l  +  i^_7  + ). 


§  7.]  GENERAL  FORMULAE.  39 

§  7,     THE   TKIGONOMETKIC   CANON. 

The  Trigonometric  Canon  is  a  set  of  tables  which  give  the 
sine,  cosine,  tangent,  cotangent,  secant  and  cosecant  for  every 
angle,  from  0  to  a  right  angle,  taken  at  regular  intervals  of  say 

10",  or  1',   or  10',  ,  as  may  be  chosen  for  the  particular 

tables.  Such  tables  contain  either  the  functions  themselves  or 
their  logarithms.  The  first  are  called  the  natural  functions,  aud 
the  others  the  logarithmic  functions. 

PROBLEM  1.    To  CONSTRUCT  A  TABLE  OF  NATURAL  SINES  AND 

COSINES,    TO    MINUTES    OF   ANGLE. 

FIRST  METHOD:  Assume  sinl9  as  differing  insensibly  from 
arc  1',  i.e.,     that  sin  1'  =  .000  290  8882, 
and  hence,     that  cos  1',  =  Vl—  sin2!',  =  .999  999  9577 ;  then : 

(1)  For  angles  0°-30°,  apply  formulae: 

sin  (0  +  6')  =  2  sin  0  cos  0'  —  sin  (0  —  6') ,  [51 

cos(0-h0')  =  2cos0cos0'  —  cos  (0  —  0')  ;  [53 

make  0=  1',  2',  3',  successively,  and  6'  —  I' constantly,  thus- 

sin  2'=  2sinl'cosl'  —  sinO' 

=  2  x  .000  290  8882  x  .999  999  9577  -  0 
=  .000581  7764  x(l-. 000  000  0423) 
=  .000  581  7764  ; 
sin  3'  =  2  sin  2'  cos  1'  —  sin  1' 

=  2  X  .000  581  7764  X  (1  —  .000  000  0423) 

-  .000  290  8882 
=  .000  872  6646. 
So,       cos2'  =  2cosl'cosl'— cosO' 

=  2  X  .999  999  9577  X  (1  -  .000  000  0423)  -  1 
=  .999  999  8308  ; 
cos  3'  =  2  cos  2'  cos  1'—  cos  1' 

=  2  X  .999  999  8308  X  (1  -  .000  000  0423) 

-.9999999577 
=  .999  999  6193*. 


40  PLANE   TRIGONOMETRY.  [II. 

(2)  For  angles  30°-45°,  substitute  30°  for  0,  and  1',  2',  3', 

successively  for  0'   in  formulae : 

sin  (0  +  0')  =  2  sin0  cos0'-  sin  (0  -  0')  [51 

=  cos0'  -  sin(0  -  0')  ;  [sin 30°  =  % 

cos  (0  -f-  0')  =  cos  (0  -  0')  -  2  sin  0  sin  0'  [54 

=  cos  (0-0')- sin  0'. 
Thus  :  sin  30°  1'     =  cos  1'—  sin  29°  59' 

=  .999  999 -  .499  75  =  .500  25  ; 

sin  30°  2'     =  cos  2'-  sin  29°  58' 

=  .999  999 -  .499  50  =  .500  50. 

So,       cos30°l'     =  cos 29°  59'-  sin  1' 

=  .866  17 -.000  29  =  .865  88; 
cos 30°  2'     =  cos  29°  58'-  sin 2'  =  .865  73. 

(3)  For  angles  45°-90°,  apply  formulae: 

sin  (45°  +  0')  =  cos  (45°  —  0')  ;  [I.  Thm.  5 

cos  (45°+  0')  =  sin  (45°-  0') . 
Thus:  sin 45°  1'       =  cos44°  59' =  .707  31  ; 

sin  45°  2'       =  cos44°  58'  =  .707  52. 
So,       cos  45°  1'       =  sin  44°  59' =.706  90; 

cos45°  2'       =  sin  44°  58'  =  .706  70. 

*  SECOND  METHOD  :  Substitute  the  value  of  0  for  0',  1',  2f, 

in  [94,  95].     Thus: 

1'-    v    =  3. 141  592  653  589  793 
180  x  60         10  800 

=  .000  290  888  208  666  ; 

...  sin  1'  =  .000  290  888  208  666  -  '00°  29°  ^  208  666  + 

=  .000  290  888  2046  ; 

.000  290  888  208  6662  , 


1f      1 
cosl=l- 


2! 
9999999577;  ' 


§  7.]  GENERAL  FORMULAE.  41 

siii  2'  =  2  X  .000  290  888  208  666 
03  ^.000290888208666* 

~TT 

=  .000  581  7764  ; 
9,_n   09  ^  .000  290  888  208  6662  , 

COS  &    —  J.  —  Z    A   —  -  —  — 

=  .999  999  8308. 

NOTE  1  .  The  process  is  evidently  very  tedious  ;  but  the  reader 
will  notice,  first,  that  it  would  be  much  shorter  if  four  or  five 
decimal  places  only  were  sought  in  the  functions  ;  and,  second, 
that  once  having  raised  the  fraction  to  the  required  powers, 
thereafter  he  has  only  to  take  simple  multiples  of  them.  At  first 
he  need  use  but  two  terms  of  the  series  ;  but  later,  when  0  is 
larger,  and  the  series  therefore  converges  less  rapidly,  more 
terms  must  be  used  ;  thus  : 

-.-,  for  30°,  0  =  -  =  .52360  nearly  ; 


=  .52360  —  .02392  +  .00033  -  .00000  -f  ..... 
=  .5  within  less  than  .00001  ; 

i.e.-,  by  the  use  of  three  terms  of  the  series,  the  sine  is  found 
correct  to  four  decimal  places,  the  same  degree  of  accuracy  as 
that  assumed  for  the  value  of  TT. 

NOTE  2.  The  results  may  be  verified  by  using  both  methods 
of  computation  ;  and  for  certain  angles  there  are  other  and  inde- 
pendent methods  : 

[63,64 


.-.  from  cos  45°,  =  Vi,  [I-  Thm.  13 

are  found  in  succession  the  sines  and  cosines  of  22°  30',  11°  15', 
5°  37'  30",  .....  . 

So,       from  cos  30°,  =  £  V3'  P-  Thm-  12 

are  found  in  succession  the  sines  and  cosines  of  15°,  7°  30', 
3°  45',  .....  . 


42                                 PLANE  TRIGONOMETRY.  [II. 

(2)     •  .  •  sin  2  0  =  2  sin  0  cos  0,  [60 

and  cos  30  =  4cos30—  3cos0,  [84 

and     v  sin  36°  =  cos  54°,  [I.  Thm.  5 

.-.  2  sin  18°  cos  18°  =  4  cos3  18°-  3  cos  18°; 


whence  are  found  in  succession  the  sines  and  cosines  of  9°,  4°  30', 
2°  15',  ..... 


(3)  Fromcos3G°  =  cos218°-  sin218°=  i(^5  +  1),     [61,  above 
and  sin36°  =  V(l-cos236°)=iV(10--2V5);   [36,  above 

are  found  the  sine  and  cosine  of  (36°—  30°),  i.e.,  of  6°,  and 
thence  in  succession  the  sine  and  cosine  of  3°,  1°  30',  45',  ..... 


(4)  Fromsm(36°+0')-sm(360-0f),=  2cos36°sin0'  [52 

=  H  V5  +  1)  sin  0',  [above 
subtract     sin(72°+0')  —  (sin  72°—  0'),  =  2cos72°sin0'  [52 

=  ^(V5  —  I)sin0'.  [above 
then,  102] 

sin(360  +  0')-sm(360-0')  =  sin(720+0')-sm(72°-0')-hsin0', 

which  is  Euler's.  formula  of  verification,  and  serves  to  test  the 
sines  of  all  angles  from  0°  to  90°,  if  to  0'  be  given  the  different 
values  from  0°  to  18°. 

PROB.  2.  To  COMPUTE  TANGENTS,  COTANGENTS,  SECANTS  AND 

COSECANTS. 

FIRST  METHOD.  For  the  tangents,  divide  the  sines  of  the  angles, 
in  order,  by  the  cosines,  each  by  each;  for  the  cotangents,  divide 
the  cosines  by  the  sines;  for  the  secants,  divide  unity  by  the  co- 
sines; for  the  cosecants,  divide  unity  by  the  sines. 

*  SECOND  METHOD.  Substitute  the  values  of  0  for  1',  2',  3',  ..... 
in  formulae  [96-99]  . 

NOTE.  These  series  converge  less  rapidly  than  those  for  sine 
and  cosine  ;  but  for  small  angles  they  may  be  used. 


§7.] 


GENEEAL  FORMULAE. 


43 


PROB.  3.    To  COMPUTE  TABLES  OF  LOGARITHMIC  FUNCTIONS  : 

FIRST  METHOD.  From  a  table  of  logarithms  of  numbers  take 
out  the  logarithms  of  the  natural  sines  and  cosines.  For  the  tan- 
gents, subtract  the  logarithmic  cosines  from  the  logarithmic  sines; 
for  the  cotangents,  subtract  the  logarithmic  sines  from  the  loga- 
rithmic cosines;  for  the  secants  and  cosecants,  subtract  the  loga- 
rithmic cosines  and  sines  from  0,  respectively. 

*  SECOND  METHOD.    For  the  sines  and  cosines,  substitute  the 

values  of  0  for  1',  2',  3', in  formulae  [100,  101]  ;  for  thetan- 

gents,  cotangents,  secants  and  cosecants,  follow  the  first  method. 

NOTE.  A  more  rapid  method,  applicable  also  to  making  tables 
of  natural  functions,  and  many  others,  is  this : 

Compute  the  logarithms  of  three,  four,  or  more  angles  at  regu- 
lar intervals,  and  lind  their  several  "  orders  of  differences  "  ; 
then,  by  the  algebraic  "  method  of  differences,"  find  the  succes- 
sive terms  of  the  series  of  logarithms, 

and  interpolate  for  other  angles  lying  between  those  of  the  series. 
Repeat  this  process  for  different  parts  of  the  table,  and  verify  by 
direct  computation.  For  safety,  four-place  tables  must  be  com- 
puted to  six  places  ;  five-place  tables  to  seven  places,  and  so  on. 

A  useful  modification  of  the  above  rule  is  this  : 

Add  the  last  difference  of  the  highest  order  to  the  last  differ- 
ence of  the  next  lower  order,  and  that  sum  to  the  last  difference 
of  the  next  lower  order,  and  so  on  till  a  term  of  the  series  is 
reached.  Thus,  in  the  example  which  follows,  the  numbers 
below  the  heavy  rules  are  got  by  successive  addition : 


Angle 

log  sine 

1st  difs. 

2d  difs. 

3d  difs. 

18° 
18°  10' 
18°  20' 
18°  30' 
18°  40' 
18°  50' 
19° 

9.  489982  4 
9.4938513 
9.4976824 
9.5014764 

38689 
38311 
37940 

-378 
371 

7 

7 
7 
7 

-364 
-357 
-350 

37576 
37219 

36869 

9.5052340 
9.5089559 
9.5126428 

44  PLANE   TRIGONOMETRY.  [II. 

*  §  8,     EXERCISES. 

If  A,  B,  c  be  any  three  plane  angles  whose  sum  is  180°  (the 
three  angles  of  a  triangle) ,  prove  that : 

1 .  tan  A  4-  tan  B  4-  tan  c  =  tan  A  tan  B  tan  c .  [80 

2.  tanj-Atan-j-B  4-  tan^B  tan^c4-tanjc  tan|-A=  1.        [80 

3.  sin2A4-sin2B  4- sin2c  =  4sinAsinB  sine.    [68,10,71,4 

4.  sinA4-sinB  4-sinc  =  4cos^-Acos^B  cos^-c.        [68,4,70 
Prove  that: 

5.  siii(2sin-1a;)=2#V(1—  O-  [60 

6.  sin  (3  sin-1  a)  =  3  a  —  4  or*.  [78 

7.  tan (2 tan'1  a)  =  2 a:  (l-x2).  [62 

8.  tan(3tan~V)=  (Sx  —  x?)  :  (1  —  3^).  [80 

11.    tan"1^  ±  tan"1?/  =  tan-1(x  ±  y)  :  (xy  ^  1) .  [43,  44 

13.   ^?r  =  tan-1l  =  tan"1!- 4- tan"1^  [43 

=  tan"1!-  4-  tan"1-!-  _j_  tan"1^  =  2  tan-1^  4-  tan"1-}-.    [80 


15.  If  (l+€cos(9)(l-ecos^)=l-£2 

then  also,  tan|0  :  tan|-^  =  «J(l  +  e)  :  ^(l  —  e).      [65 

16.  IfO.O'  and  6"  be  any  three  angles,  prove  that 

_ 

~  :*V* 


,,   f  9 
17.    If  6  and  0'  be  any  two  angles  ,  and  n  any  integer,  prove  that 


[sm0+sin(<9-|-<9')+sm(0-j-2<9f)-{- 

=  cos  (0  -  J0')  -  cos  (0  +^—!>6f)  ,  [54 

[cos0+cos(0+0')+cos(0+2<9')_f-  .....  +cos(0+?i^l)0'].2sin40' 

=  sm(0  +  '^=l6')-sm(0-±0').  [52 

18.  From  the  results  of  Ex.  17,  prove  that 


§  8.]  GENERAL  FORMULAE.  45 


\        n  \        n 

wherein  n  is  any  positive  integer. 

19.  In  the  results  of  Ex.  18,  make  n  =  3,  and  prove  that 

sin  9  +  sin  (60°  -  0)  -  sin  (60°+  0)  =  0  ; 
cos  0  —  cos  (60°  -  0)  -  008(60°+  0)  =  0. 

20.  In  the  results  of  Ex.  18,  make  n  =  5,  and  prove  that 
sm0+sin(720+0)+sin(360-0)-sin(360+0)-sin(72°-0)==0, 
cos0+cos(720+0)-cos(360-0)-cos(360+0)+cos(72°-0)=0, 
and  from  the  first  of  these  two,  get  Euler's  formula.  [102 

21.  In  the  results  of  Ex.  18,  make  n  —  9  and  15,  in  succes- 
sion, and  thence  find  other  formulae  of  verification. 

Show  that  the  formula  found  when  n  =  9  verifies  the  sines  of 
all  angles  in  the  quadrant,  if  to  0  be  given  values  from  0°  to  10°. 

22.  If  6  be  an}'  angle  and  0'  be  an  infinitesimal  angle,  the  in- 
crement of  0,  prove  that 

inc2sin0  =  -(2sin£0')2sm(0  +  0'),  [69 

inc2cos0  =  -  (2  sin  £0')2  cos  (0  +  0')  ,  [71 

inc4sin  0  =      (2sin|0')4sin  (0  +  20'), 

inc4cos0  =      (2sn40')4cos(0  +  20'), 

Wherein  inc2sin  0  stands  for  the  increment  of  the  increment  of  sin  0, 
i.e.,  for  «  [sin  (0  +  20')  -  sin(0  +  0')]  -  [sin(0  -f-  0')  —  sin0], 
or  sin  (0  -f  2  0')  —  2  sin  (0  +  0')  +  sin  0, 

and  inc40  stands  for  inc  inc  inc  inc  sin  0,  i.e.,  for 
sin(0+40')-4sin(0+30')  +  6sin(0+20')-4sin(0+0')  +  sin0 
and  so  on.  [Alg.,  Meth.  Dif. 

23.  Compute  the  sines  and  cosines 

of  22°  30'  and  G7°  30',     11°  15'  and  78°  45',  .....  ; 

of  15°        and  75°,  7°  30'  and  82°  30',  .....  ; 

of    6°        and  84°,  3°        and  87°,         .....  ; 

of    9°        and  81°,  4°  30'  and  85°  30',  ...... 

[Prob.  1,  note  2 

24.  From  the  logarithmic  sines  of  18°,  18°  10',  18°  20',  ..... 
find  the  several  orders  of  differences,  and  thence,  by  interpola- 
tion, find  the  logarithmic  sines  of  18°  1',  18°  2',  18°  3',  18°  4',  ..... 


46 


PLANE  TRIGONOMETRY. 


Fill. 


HI.     SOLUTION  OF  PLANE  TRIANGLES. 

§  1,     GENERAL   PROPEETIES   OF   PLANE   TRIANGLES. 

THEOREM  1.     In  any  plane  triangle  the  sides  are  proportional 
to  the  sines  of  the  opposite  angles. 


B  A 


Let  ABC  be  any  triangle  ;  a,  &,  c  the  sides  opposite  the  angles 
A,  B,  c  respectively  ;  then  will : 

103]  a •:  b  =  siiiA  :  sinB  ; 

b  :  c  =  sinB  :  sine  ; 

c  :  a  =  sin  c  :  sin  A. 
For,  draw  DC  _L  AB  ;  then 

DC  =  AC  sin  A  =  b  sin  A, 
and  DC  =  BCsiiiB  =  asinB, 


[I.  §  7,  note 


So, 
and 


a  :  b  =  sin  A  :  sinB. 
b:c  =  sinB:  sine, 
era  =  sin  c  :  sin  A. 


[Thm.  prop'n 


Q.E.  D, 


NOTE  1.     This  theorem  may  be  stated  more  symmetrically 
thus: 

a :  b  :  c  =  sin  A  :  sin  B  :  sin  c ; 
or  thus : 

a  b  c 

sin  A     sin  B     sin  c 


§  1.]  SOLUTION  OF  PLANE  TRIANGLES.  47 

COR.    In  any  plane  triangle  ABC  : 

sinB        sine 

,       csiiiB      asinB 

b  =  — =  — ; 

sine        sin  A 

__asinc_  b  sin  c  . 

sin  A       shiB 
105]  BinA  =  5L«EB  =  £8inc; 

b  c 

bs'mc      b  sin  A 


sins  = 


sin  c  = 


c  a 

csinA      csiiiB 


NOTE  2.  The  expressions  -  ,  .....  are  equal  to  the  diameter 
sin  A 

of  the  circumscribed  circle,  as  appears  later,  PROB.  4. 

NOTE  3.  All  the  angles  of  a  triangle  are  positive  angles,  and 
the  sides  of  the  triangle  arc  positive  lines  when  taken  with  refer- 
ence to  the  angles  included  b}"  them. 

The  ordinatc  of  the  vertex,  with  reference  to  the  base,  is  there- 
fore alwatys  positive  ; 

but  the  abscissa  ma}T  be  either  positive  or  negative,  .according  as 
it  is  measured  in  a  positive  or  negative  direction  from  the 
origin  used. 

THM.  2.    In  any  plane  triangle,  the  sum  of  any  two  sides  'is  to 
their  difference  as  the  tangent  of  half  the  sum  of  the  two  opposite 
angles  is  to  the  tangent  of  half  their  difference  : 
106]  i.e.,  (a  +  6)  :  (a~  &)  =  tan  J(A  +  B)  :tan£(A~B)  ; 


(c  +  a)  :  (c  ~a)  =  tan£(c  -f  A)  :  tan  £  (c  ~  A)  . 
For,  •.*  a  :  b  =  sin  A  :  sinB,  [Thin.  1 

.-.  (a  +  b)  :  (a  —  b)  =  (sin  A  -f  SIHB)  :  (sinA  —  sins)  ; 

[Thm.  prop'n 

and  *.•  (sin  A  +  sin  B):  (sin  A—  siiiB)  =  tan£(A+B)  :  tan£(A—  B)  ; 

[72 
.-.  (a+l)  :  («-&)  =  tan  J(A  +  B)  :tan'(A-is). 


48 


PLANE   TBIGONOMETBY. 


[III. 


But*. 'the  greater  side  of  a  triangle  lies  opposite  the  greater 
angle,  \_geom. 

.  • .  when  a  >  6,  then  also  A  >  B, 

and       (a  +  &)  :  (a  -  b)  =  tan|-(A +  B)  :tan|(A  — B). 
So,       when  b  >a,  then  also  B  >  A, 
and       (a  +  b)  :  (b  —  a)  =  tan£(A  +  B)  :tanJ-(B— A)  ; 
.  • .  (a  -f-  b)  :  (a  ~  b)-=  tan  -£(A  +  B)  :  tan  £  (A  ~  B)  . 
So,        (b  -f-  c)  :  (b  ~  c)  =  tan  i  (B  +  c)  :  tan  £(B  ~  c) , 
and       (c +a)  :  (c  ~a)  =  tanJ-(c  +  A)  :  tan|(c  ~  A).      Q.E.  D. 
COR.    In  any  plane  triangle  ABC  : 

107]     tan£(A~B)  =  ^-? 

a  4-  / 

1A  6   ~'   ( 


tan |( 


,A)-£^f 

c  4-  a 

THM.  3.     J;i  any  plane  triangle  ABC  : 
108] 


;   COSC  = 


2a6 


A 

c      D 

B    A 

B 

For, 

•.• 

a2=&2  +  c2_2 

and 

AD  =  b  COS  A  ; 

D        P 


a2  =  62  +  c2  —  2  6c  cos  A, 


[I.  §  7,  note 


and 
So, 
and 


COSA  = 


COSB  = 


cosc  = 


2  be 


2ca 


Q.E.D. 


§  1.]  SOLUTION  OF  PLANE  TUTAXGLES.  49 

TIIM.  4.    In  any  plane  triangle  ABC  : 

1091  sin4-A  =  -v/—  — > 

J  \  be 

-c)(s-«] 


ca 
(s-a)(s-b) 


wherein  s  stands  for  £(a  +  b  +  c)  ,  or  half  the  perimeter. 

For,    v      2sin2£A  =  1  —  cos  A  [63 

_a2  ^ 


26c 

(a  —  6  +  c)  (a  +  &  —  c) 
2  be 


So,  8inJB  =  J^-c)(^a>; 


and  Blnjc  =          --.  Q.B.D 


THM.  5.    In  any  plane  triangle  ABC 

1101  cosJ-A=      ls(s-a). 

.     "Xl      be 

ls(s-b)_ 

~  \      ca 


50  PLANE  TRIGONOMETRY.  [III. 

For,     v     2cos2|A  =  l+cosA  [64 


c-a 

2  be 


2  be 


2  be 

_4s(s  —  a). 
2bc 


/*(*-&). 

80,  COS^B        =-vf—  -  -  -1 

II  CCv 

^c     =  \ 


and  cos 


CCv 

" 


THM.  6.    Jw  an?/  plane  triangle  ABC 
111] 


s(s-c) 
For,  tan^A  =  sin^A:  COS^-A  [35 


s  (s  -  a) 

\(s  - 
So, 


and  tanc=. 


NOTE.  Since  no  angle  is  greater  than  180°,  therefore  no  half- 
angle  is  greater  than  90°,  and  the  radicals  of  Thins.  4,  5,  and  6 
are  all  positive.  [I.  Thm.  3 


§  2.]  SOLUTION  OF  PLANE  TRIANGLES.  51 

§  2.       SOLUTION    OF    BIGHT   TRIANGLES. 

PROBLEM  1 .    To  SOLVE  A  RIGHT  TRIANGLE  : 

Let  OAF  be  an}r  right  triangle  ;  o,  the  base- angle  ; 
A,  the  right  angle ;  x,  y,  and  r,  the  base,  perpen- 
dicular and  hypothenuse. 

CASE  1.  Given  the  hypothenuse  and  an  acute 
angle,  for  example  r  and  o ;  then  : 

P=90°-o; 


or    =  rsinp; 
2/  =  rsino,     or    =  rcosp,    or    =  #tano.  [I.  §  7 

CASE  2.   Given  a  side  and  an  acute  angle,  for  example  x  and  o  ; 

then: 

p=90°-o; 

r  =  x  :  cos  o,     or    =  x  :  sin  p  ; 

y  =  xtano,       or    =  rsino,     or    =  rcosp.         [I.  §  7 

CASE  3.  Given  the  hypotlienuse  and  a  side,  for  example  r  and  x  ; 

then  : 

coso  =  x  :  r,     whence  o  is  found  ; 

P  =  90°—  o; 

2/  =  rsino,     or    =rcosp,     or    =  #tano,     [I.  §  7 
or      =  Vr2—  x2  =  V  (r  +  ic)  (r  —  x)  .  [geom. 


CASE  4,    Given  the  two  sides  about  the  right  angle,  then  : 
tan  o  =  y  :  x,     whence  o  is  found  ; 


P=90°-o; 


r  =  #:coso,     or    =  a;:  ship,     or    =y:sino, 
or      =  y :  cos  p,  [I.  §  7 

or      =  Va^H-  y2.  •  \_geom. 

NOTE  1.    To  test  the  correctness  of  the  work,  various  checks 
may  be  applied  : 

(1)  Compute  the  part  by  some  other  process. 

(2)  Substitute  the  three  computed  parts  in  some  one  of  the 
six  equations  which  express  the  definitions  of  the  trigonometric 
functions  [I.  §  7];  if  this  equation  is  satisfied  the  parts  are  right. 


52  PLANE  TEIGONOMETKY.  [III. 

functions  [I.  §  7]  ,  or  of  the  six  equations  which  result  therefrom 
[I.  §  7,  note]  ;  if  this  equation  is  satisfied  the  parts  are  right. 

(3)   Take  the  three  given  parts  and  one  computed  part, 
or  two  given  parts  and  two  computed  parts, 
or  one  given  part  and  the  three  computed  parts, 
and  substitute  them  in  some  one  of  the  formulae  of  Thms.  1-6  ; 
if  this  formula  is  satisfied,  the  work  is  correct,  for  the  part  or 
parts  tested. 

*NOTE  2.  When  the  solution  involves  angles  near  to  0°,  90° 
or  180°,  care  must  be  used  in  the  selection  of  formulae  ;  for,  of 
angles  near  90°,  those  which  differ  very  considerably  have  nearly 
the  same  sine,  and  cannot  therefore  be  determined  with  precision 
from  the  table  of  sines,  thus  : 

the  sines  of  all  angles  from  89°  50'  to  90°,  inclusive,  differ  from 
1  by  less  than  .000005,  and  the  sines  of  all  angles  from  89°  49' 
to  89°  42',  inclusive,  differ  from  .99999  by  less  than  .000005  ; 
but,  of  tangents,  that  of  89°  is  57.2900  ;   that  of  89°  20'  is 
85.9398  ;  that  of  89°  40'  is  171.885  ;  that  of  90°  is  infinity  ; 
whereby  it  appears  that  the  tangents  of  angles  near  90°  not  only 
increase  very  fast,  but  that  they  also  increase  faster  and  faster. 

So,  of  angles  near  0°  and  180°,  the  natural  cosines  change  very 
slowl}',  and  the  natural  cotangents  ver}7  fast. 

Of  angles  near  0°  and  180°,  the  logarithmic  sines,  tangents 
and  cotangents  change  very  fast,  and  at  rapidly  varying  rates, 
and  the  logarithmic  cosines  very  slowly. 

So,  of  angles  near  90°,  the  logarithmic  cosines,  cotangents  and 
tangents  change  very  fast,  and  at  rapidly  varying  rates',  and  the 
logarithmic  sines  very  slowly. 

To  avoid  these  angles,  use  the  following  formulae  : 

(1)  If  x  be  very  small  compared  with  r,  and  therefore  o  be 
nearly  90°,  use  : 

1  7*       -  />* 

112]  sin£o  =  --r; 


113] 


§  2.]  SOLUTION   OF  PLANE   TRIANGLES. 

114] 


For,  v     coso  =  ->  [I-  §  7 

r 

.'.     2sin2^o,     =1— coso,     =  1 — ,  [63 

/•W  At         I  /V» 

and  2cos2^o,     =l+coso,     =l-f-      =- ,  [64 

.'.     etc.  Q.  E.  D. 

(2)  If  x  be  nearly  equal  to  r,  and  therefore  o  be  nearly  0, 
compute  ?/,  =  V  (r  +  x)  (r  —  x) ,  and  use  : 

115]         sm^p: 


116] 

\    2r 

117] 


(3)  If  #  be  very  small  as  to  y,  and  therefore  o  }}e  nearly  90°^ 
compute  r,  =Vit>2+2/2,  then  use  either  of  the  formulae  [112- 
114]  ;  if  ?/  be  very  small  as  to  #,  use  [115-117]. 

*NOTE  3.  If  o  be  very  small,  then  special  tables  may  be  used 
which  give  the  angle  in  seconds,  the  logarithmic  sine  and  tan- 
gent, and  the  logarithms  of  the  ratios  of  the  sine  and  tangent  to 
the  number  of  seconds  in  the  angle.  These  ratios  are  nearly 
constant,  and  their  differences  are  very  nearly  proportional  to 
the  differences  of  the  corresponding  angles.  The  formulae  are  : 

118]  o  in  seconds  ___  EL2  __  =  I  ; 

o  in  seconds      r 

119]       and     o  in  seconds  .  _  tano  _  =  .?. 

o  in  seconds     x 

So,  if  p  be  ver}'  small. 

These  special  tables  give  the  functions  and  angles  with  more 
accuracy  than  the  ordinary  tables  give  functions  and  angles  in 
any  part  of  the  quadrant. 


54 


PLANE   TRIGONOMETRY. 


[Ill 


§  3,     SOLUTION    OF    OBLIQUE    TRIANGLES. 
PROB.  2.    To  SOLVE  AN  OBLIQUE  TRIANGLE. 

jf 


FIRST  METHOD.    By  means  of  right  triangles. 

Let  ABC  be  any  oblique  triangle ;  and  a,  6,  c  the  sides  oppo- 
site the  angles  A,  B,  c,  respectively. 

From  c  draw  DC  JL  AB,  and  let  y  stand  for  DC,  x'  for  AD,  x"  for 
DB,  c'  for  ZACD,  and  c"  for  ZDCB. 

CASE  1.  Given  two  angles  and  a  side;  for  example,  A,  c  and 
6  ;  then : 

B  =  180°-(A  +  c);  [geom. 

In  rt.  A  ACD,  b  and  /.  CAD  are  known,  whence  y  and  x1 
are  found ; 

In  rt.  A  BCD,  y  and  /.  CBD  are  known,  whence  a  and  x" 
are  found ; 

c  =  x'+x". 

NOTE  1.  The  reader  must  carefully  distinguish  between  the 
signs  of  the  sides  and  angles  of  a  triangle,  when  taken  with  refer- 
ence to  the  triangle  itself,  and  when  taken  with  "reference  only  to 
some  initial  direction  ;  thus  : 

The  parts  of  the  triangle  BCD,  when  taken  with  reference  to  the 
triangle  BCD,  are  all  positive  ; 

but,  with  reference  to  AB,  as  initial  direction,  the  line  DB  is  posi- 
tive or  negative  according  as  it  is  measured  to  the  right  or  to 
the  left  from  D. 

So,  with  reference  to  ordinary  positive  rotation,  the  angle  DCB  is 
positive  or  negative  according  as  CB  swings  to  the  right  or  to 
the  left  from  CD  as  the  reader  looks  at  the  diagram. 


§3.]  SOLUTION  OF   PLANE   TKIANGLES.  55 

NOTE  2.  x'  and  c'  are  positive  or  negative,  with  reference  to 
AB  and  to  ordinary  positive  rotation,  according  as  A  is  acute  or 
obtuse  ;  and  x"  and  c"  are  positive  or  negative,  according  as  B 
is  acute  or  obtuse. 

If  A  +  c  =  or  >  180°,  there  is  no  triangle  ; 
if  A  -f-  c  <  180,  there  is  always  one  triangle,  and  but  one. 
The  parts  are  determined  without  ambiguity  from  the  formulae. 

CASE  2.  Given  two  sides  and  the  included  angle;  for  example, 
&,  c  and  A  ;  then  : 

In  rt.  A  ACD,  b  and  ZCAD  are  known,  whence  y,  x'  and 

c'  are  found  ; 
x"=c-x'; 
In  rt.  A  BCD,  y  and  x"  are  known,  whence  Z  CBD  and 

c"  are  found  ; 
c  =  c'+c". 
B  =  ZCBD     or    =180°—  CBD. 

NOTE,  x'  and  c'  are  positive  or  negative,  with  reference  to  AB 
and  to  ordinary  positive  rotation,  according  as  A  is  acute  or 
obtuse, 

and  x"  and  c"  are  positive  or  negative,  and  B  is  acute  or  obtuse, 
according  as  c>  or  <xf. 
There  is  alwa}*s  one  triangle  and  but  one  ; 
the  parts  are  determined  without  ambiguity  from  the  formulae. 

CASE  3.    Given  the  three  sides  ;  then: 

and     62  =  2/2-fo;'2,  [geom. 


.-.     (a  +  b)(a  —  b)  =  (x"  +  x',  or  c)  (x"  —  x')  ,  whence  x"  —  x' 

is  found, 
and      x1  =  ic  -  %(x"-  x')  ,         x"=  $c  +  J(a?"-  x1)  . 

In  rt.  A  ACD,  b  and  x'  are  known,  whence  Z.  CAD  is  found  ; 
In  rt.  A  BCD.  a  and  x"  are  known,  whence  Z  CBD  is  found  ; 
c=180°-(A-f  B). 


56 


PLANE   TRIGONOMETRY. 


[III. 


NOTE.  A  and  B  are  acute  or  obtuse,  i.e.,  equal  or  supplemen- 
tary to  CAD  and  CBD,  according  as  x1  and  x"  are  respectively  posi- 
tive or  negative  with  reference  to  AB. 

There  is  no  triangle  if  either  side  equals  or  exceeds  the  sum  of 
the  other  two  ;  otherwise,  there  is  one  triangle,  and  but  one. 
The  parts  are  determined  without  ambiguity  from  the  formulae. 

CASE  4.  Given  two  sides  and  an  angle  opposite  one  of  them; 
for  example,  a,  b  and  A. 

In  rt.  A  ACD,  b  and  /.  CAD  are  known,  whence  y,  x'  and 

c'  are  found  ; 

In  rt.  A  BCD,  a  and  y  are  known,  whence  a?",  Z  CBD  and 
c"  are  found  ; 


B  =  ZCBD    or    =180°—  CBD,  according  as  c"  is  added 

to,  or  subtracted  from,  c'. 
NOTE.     There  may  be  one  triangle,  two  triangles,  or  none. 


(1)  If  A  is  acute  and  a  <  5,  but  a  >  ?/,  there  are  two  triangles  ; 

(2)  if  A  is  acute  and  a  <  6,  but  a  =  ?/,  there  is  one  triangle  ; 

(3)  if  A  is  acute  and  a  <  6,  but  a  <  y,  there  is  no  triangle. 

(4)  If  A  is  acute  and  a  =  &,  there  is  one  triangle  ; 

(5)  if  A  is  acute  and  a  >  6,  there  is  one  triangle. 

(6)  If  A  is  right  or  obtuse  and  a  >  6,  there  is  one  triangle  ; 

(7)  if  A  is  right  or  obtuse  and  a  =  &,  there  is  no  triangle ; 

(8)  if  A  is  right  or  obtuse  and  a  <  6,  there  is  no  triangle. 


§3.]-  SOLUTION   OF  PLANE   TRIANGLES.  57 

For  v     c"  is  found  from  its  cosine,  =  y :  a, 

.'.    when  a  <6,  c"  and  x"  may  be  either  positive  or  nega- 
tive ;  andB  =  CBD    or    =180°— CBD.  [2 
But,  when  a  =  or  >  6,  the  negative  values  of  x"  and  c"  are  in- 
admissible ;  for,  then  c,  =  x'+'x",  and  c,  =  c'+"c",  are  both  0 
or  negative,  which  is  absurd. 

SECOND  METHOD.     By  means  of  the  general  properties. 
CASE  1.     Given  two  angles  and  a  side;  for  example,  A,  B  and 
c,  then : 

c  =  1 80°  —  (  A  +  B)  ,  [geom. 

(*  (* 

a  = sin  A,  b=  —  —sins.  [104 

sin  c  sin  c 

CJiecTc:   See  formulae  [120,  121]. 

NOTE.  The  formulae  give  one  value,  and  but  one,  for  each  part. 

CASE  2.    Given  two  sides  and  the  included  angle;  for  example, 
a,  b  and  c  ;  then  : 

.   ^^    7, 

tan£(A~B)  =  —    —  cotjc,  whence  -J(A~B)  is  found, 

[107 

•J-(A  +  B)  -f-  ^(A  ~  B)  =  the  greater  of  the  two  angles  ; 

•J(A  -+-  B)  —  ^(A  ~  B)  =  the  less        of  the  two  angles. 

c  =  -^— sine.  [104 

sin  A 

Check :   b  :  sin  B  =  a  :  sin  A. 

NOTE.  The  formulae  give  one  value,  and  but  one,  for  each  part. 

CASE  3.     Given  the  three  sides:  a,  5,  c. 

Apply  the  formulae  of  Thms.  3-6. 

Check:  A  +  B  +  c  =  180°. 

NOTE  1.    The  formulae  give  one  value,  and  but  one,  for  each 
part. 

NOTE  2.     Of  these  formulae,  those  of  Thm.  3  are  only  useful 
when  the  computation  is  by  natural  functions  ; 
those  of  Thm.  4  use  nine  different  logarithms, 
those  of  Thin.  ">  use  ten  different  logarithms, 
those  of  Thm.  6  use  only  seven  different  logarithms, 
for  the  computation  of  all  the  angles.     The  formulae  of  Thm.  6 


58  PLANE  TRIGONOMETRY.  [III. 

are  therefore  generally  to  be  preferred  ;  they  may  be  put  in  the 
form: 


—  L-     l(-a)(-6)(»-c) 

s—  a\  s 


(s  —  a)  (s  —  b)(s  —  c) 

5 

(—  a)  («-&)(«  -J1  . 

—  c  \  s 

wherein  the  second  factor  of  the  right  member  is  the  same,  and 
may  be  computed  but  once,  for  all.  It  will  appear  later,  PKOB.  4, 
that  this  factor  is  the  radius  of  the  inscribed  circle. 

CASE  4.    Given  two  sides  and  an  angle  opposite  one  of  them; 
for  example,  a,  b  and  A  ;  then  : 

shin  =  -  sin  A,  whence  B  is  found.  [104 

Ch 

c  =  180°  -  (A  +  B)  ,  [geom. 

sine.  [104 


a 


smA 

Clieck:   See  formulae  [120,  121]. 

NOTE  1 .    The  formulae  leave  the  parts  in  doubt ; 
for  the  same  value  of  SIUB  belongs  to  two  angles,  which  are 
supplements  of  each  other  $ 

so  that,  in  general,  B  may  be  an  acute  or  an  obtuse  angle ; 
whence  two  values  each  for  c  and  c,  and  two  triangles. 
But  this  is  limited  b}*  the  conditions,  that  "  the  greater  side  of  a 
triangle  lies  opposite  the  greater  angle,"  and  that  "a  triangle 
can  have  but  one  obtuse  angle,  and  no  side  0  or  negative." 
If,  then,  a  =  or  >  6,  B  cannot  be  obtuse  ; 
and  if  A  is  obtuse,  B  is  acute. 

Moreover,  the  shortest  length  possible  for  a  is  the  perpendicular  DC  ; 
for    •••     6sinA  =  asiiiB  =  DC, 

.*.     if  a  <  DC,  then  sins  >  1, which  is  impossible. 

NOTE  2.  The  same  care  must  be  taken  in  solving  oblique  tri- 
angles as  in  solving  right  triangles,  to  avoid  the  use  of  angles 
near  0°,  90°  or  180°,  unless  the  special . table  mentioned  under 
right  triangles  is  used. 


*.]  SOLUTION  OF  PLANE  TRIANGLES.  59 

The  following  formulae  are  useful ;  the  reader  ma}^  prove. 


121] 
122] 
123] 
124] 
1251 

c 

c 
s 

sin^c 
sin  ^  (A  ~  B) 

COSi(A  +  B)' 

sin|(A~B)  . 

cos^c 

COS  £  A  COS  £  B  . 

sin  ^  (A  -}-  B)  ' 

c 
s  —  c 
c 
s-b 

sin^c 

sin  1  c 
sin  J  A  COS!B  . 

c 
s  —  a 

cos  •>  c 
cos  •}  A  sin  •>  B 

§4.      THE   AREA   OF   A   TRIANGLE. 
PROB.  3.    To  FIND  THE  AREA  OP  A  TRIANGLE,  AND  THE  LENGTH 

OF   THE   PERPENDICULARS   FROM    THE   VERTICES    TO    THE    OPPOSITE 
SIDES  : 

Let  ABC  be  aiw  triangle,  and  let  K  stand 
for  its  area,  and  pa,  pb,  pc,  for  the  perpen- 
diculars on  a,  6,  c,  respectively. 

CASE  1.  Given  two  sides  and  the  included 
angle;  for  example,  6,  c,  and  A. 

(1)  For  the  area,  multiply  half  the  product 

of  any  two  sides  by  the  sine  of  the  included    A 
angle. 

(2)  For  the  perpendicular  upon  either  given  side,  multiply  the 
adjacent  side  by  the  sine  of  the  included  angle. 

For,  draw  DC_L  AB  ;  then, 

V      K  =  ^DC.AB  =  |J?OC;  [geom. 

and      v     _pc=&shiA;  [I.  §  7,  note 

120]     .'.      K  =  ^6csinA. 
So,  K=Tj-casinB; 

and  K 


60  PLANE   TRIGONOMETRY.  [Ill 

CASE  2.    Given  the  angles  and  one  side;  for  example,  c  : 

(1)  For  the  area,  multiply  half  the  square  of  any  side  by  the 
sines  of  the  adjacent  angles,  and  divide  the  product  by  the  sine 
of  the  opposite  angle. 

(2)  For  the  perpendicular  ,  multiply  the  side  by  the  sines  of  the 
adjacent  angles,  and  divide  the  product  by  the  sine  of  the  opposite 
angle. 

For,        v    K  =  ^6csinA,  [126 


and        v     b  =  -,  [104 

sine 

127]       .'.    K= 


sine 

4  a2  sin  B  sine  1  &2  shi  c  shi  ^ 

So,  K  =  —   —  -  -  —  ,       and     K  = 


128 


sin  A  smB 

2K      a  sin  B  sine 


a  smA 


b  sin  c  sin  A  c  sin  A  sin  B 

So,  pb=  --  :  —    —5         and    pc  — 


smB  sine 

CASE  3.    Given  the  three  sides,  a,  b,  c: 

(1)  For  the  area,  from  half  the  sum  of  the  sides  subtract  each 
side  separately,  multiply  the  continued  product  of  these  remainders 
by  the  half  sum, -and  take  the  square  root  of  the  product. 

(2)  For  tfie  perpendicular,  divide  twice  the  root  above  found 
by  the  side  on  which  the  perpendicular  falls. 

For,     v       K  =  ia6sinc;  [126 

and      v  sine  =  2sin^ccos^c  [60 

\(s~5).J*(g7c)          [109,  no 

ab  \      ab 

2  ", 


=  2. 


129]      .'.     K  =  Vs(s  —  a)  (s—b)  (s  —  c), 


1301          •  =  = 

J  la      a 

2K  2K 

So,  ^^-5"'  and      ^«  =  ~ 


§  5.]  SOLUTION   OF   PLANE   TKI  ANGLES. 

NOTE.    The  reader  may  prove  the  cheek-formula 


61 


§5,      INSCRIBED,  ESCRIBED  AND    CIRCUMSCRIBED    CIRCLES. 
PROB.  4.     To  FIND  THE  RADII  OF  THE  CIRCLES  INSCRIBED  IN, 

ESCRIBED  AND  CIRCUMSCRIBED  ABOUT,  ANY  TRIANGLE. 

(1)  For  the  radius  of  the  inscribed,  circle,  divide  the  area  by 
half  the  perimeter. 

(2)  For  the  radius  of  an  escribed  circle,  divide  the  area  by 
half  the  perimeter,  less  the  side  beyond  which  the  circle  lies. 

(3)  For  the  radius  of  the  circumscribed  circle,  divide  half  of 
either  side  by  the  sine  of  the  opposite  angle. 


/' 


For,  let  ABC  be  any  triangle,  and  let  r  stand  for  the  radius  of 
the  inscribed  circle,  and  r',  r",  r'",  for  the  radii  of  the  escribed 
circles  whose  centers  are  o',  o",  o'",  respectively ; 
let  R  stand  for  the  radius  of  the  circumscribed  circle ;  then : 


62 

PLANE   TRIGONOMETRY. 

[III. 

(1)  •.• 

K  =  Jr(a  +  b  +  c)  =  ?*£• 

[greom. 

132]     .-. 

,  _  K         l(s  —  ct)  (s  —  6)(s  —  c) 

Q.  E.  D. 

s       \                   s 

(2)   ..- 

K  =  ir'(-«  +  &+c)  =  ?''(s-a)' 

=  J  r"  (a  -  6  +  c)  =  r"  («-&), 

=  ir'"(a  +  6-c)  =  r'"(s-c); 

[geom. 

1331        • 

K                            K                             K 

Q.  E.  D. 

NOTE.   The  reader  m&y  prove  the  check-formulae 


134] 


_   *.         o-'        /)-"      1'"' 


and 


136]     .-.       R  =  -^ 
J  sin  A 


So, 


135]        K2  =  r  .  rr .  r". 

(3)  About  ABC  circumscribe  a  circle  and 
draw  CA'  a  diameter ;  join  A'B,  then 

A  =  A',  and  Z  A'BC  is  a  rt.  Z,  [geom. 

[I.  §7. 


CA'=^ 


sm  A'     Bin  A 


sin  B      sm  c 

§  6,      EXERCISES. 


Q.E.D. 


Solve  the  right  triangles  [both  by  use  of  natural  functions  and 
by  use  of  logarithmic  functions] ,      given  : 

p/  I.  r  =  36.3,  o  =  50°. 

2.  x  =29.28,  p  =  37°!2'. 

3.  r  =  125,  y=l05. 

4.  a;=29.275,  y  =  39.07. 

5.  r  =  37.09,  y=.379. 

6.  r  =  !3ll,  o  =  89°l8f. 


SOLUTION   OF  PLANE   TRIANGLES. 


63 


Solve  the  oblique  triangles  [both  methods],      given: 


7.  a  =  25.3, 

8.  A  =  34°, 

9.  a=127, 

10.  a  =18, 

11.  a  =10, 

12.  a=16, 
!:•>.    a  =  20, 
14.    a  =24, 

a  =  24, 
a  =  20, 
a=16, 
a=127, 


15. 
16. 
17. 

18. 


19.    a  =2000, 


6=136, 
B  =  95°, 
6  =  64.9, 
6  =  20, 
6  =  20, 
6  =  20, 
6=20, 
6=20, 
6  =  20, 
6  =  20, 
6  =  20, 
6  =  254, 
6=1999, 


c  =  98°  15'. 
c  =  13.89. 
c=  152.16. 
A  =  55°  2.4'. 
A  =  30°. 
A  =  86°  40'. 
A  =  47°  9'. 
A  =  37°  36'. 
A  =120°. 
A  =  135°. 
A  =150°. 
c  =  380. 
A  =  91°. 


D          15 


Find  the  areas  of  the  triangles,  and  the 
three  perpendiculars,  jpfl,  p6,  pc,      given : 

20.  0  =12.5,         6  =  25,  c  =  36°. 

21.  A  =  37°  18',     B  =  92°18',    c  =  39.5. 

22.  a  =  29. 7,         6  =  6.238,      c  =  34.21 

Find  the  radii  of  the  inscribed,  escribed  and  circumscribed 
circles,      given : 

23.  a  =  12. 7,  6  =  22.8,  c  =  33.9. 

24.  A  =  64°  19' 8",  B  =  100°  2' 27",    c  =  51.25. 

25.  a  =136,  6  =  95.2,  c=ll°37'. 

26.  In  surveying,  the  bearing  of  a  point  is  the  angle  which  its 
direction  makes  with  the  north  and  south  line  through  the  point 
of  observation  ;  its  latitude  is  its  distance  north  or  south,  and  its 
departure  is  its  distance  east  or  west,  from  the  datum-point. 

If  a  surveyor,  starting  from  A, 
run       N.  22°  37'  E.,  3.37   chs.  to  B  ; 
thence  N.  80°  24'  E.,  3.81   chs.  to  c  ; 
thence  s.  41°  12'  E.,  5.29   chs.  to  D  ; 
thence  s.  62°  45'  w.,  6.22£chs.  to  E  ; 
find  the  latitude  and  departure  respectively 
of  B,  c,  D,  E,  from  A  ;  and  find  the  bearing 
and  distance  of  A  from  E. 


PLANE   TRIGONOMETRY. 


[III. 


27.  Divide  the  field  above  given  into  triangles  and  trapezoids, 
by  means  of  parallels  of  latitude  (east  and  west  lines)  through 
the  corners,  and  thence  find  its  area. 

28.  At  120  feet  distance  from  the  foot  of  a  steeple  standing 
on  a  plane,  the  angle  of  elevation  of  the  top  is  60°  30f ;  find  the 
height. 

29.  From  the  top  of  a  rock  326  feet  above  the  sea  the  angle 
of  depression  of  a  ship's  hull  is  24°  ;  find  the  distance  of  the  ship. 

30.  A  ladder  29£  feet  long  standing  in  the  street  just  reaches 
a  window  24  feet  high  on  one  side  of  the  street,  and  21  feet  high 
on  the  other  side  ;  how  wide  is  the  street? 

31.  What  is  the  dip  of  the  horizon  from  the  top  of  a  moun- 
tain 2f  miles  high,  the  earth's  mean  radius  being  3956  miles  ? 

32.  From  the  top  of  a  mountain  1J  miles  high,  the  dip  of  the 
horizon  is  1°36'52"  ;  find  the  earth's  diameter. 

33.  Given  the  earth's  mean  radius  3956  miles,  and  the  angle 
which  this  radius  subtends  at  the  sun  8". 81  ;  find  the  distance  of 
the  earth  from  the  sun. 


Fig.  1. 


Fig.  2. 


Fig.  3. 


34.  What  is  the  distance  across  a  river,  when  the  base  AB 
=  475  ft.,  ZA  =  90°,  ZB  =  57°  13' 20"?     (Fig.  1.) 

35.  What  is  the  distance  CD?     (Fig.  2.)     Given:   the  base 
AB  =  131-J  yds.,  Z.  BAD  =  50°,  Z.  BAC  =  85°  15',  ZDBC  =  38° 43', 
/.  DBA  =  94°  13'.     Prove  the  work  by  making  two  distinct  com- 
putations from  the  data. 

36.  What  is  the  distance  AB?    (Fig.  3.)     Given  :  CA  =  131  ft. 
5  in.,  BC  =  109  ft.  3  in.,  and  /.  c  =  98°  34'.     Prove  the  work. 


*S/TY 

§  C.]  SOLUTION   OF   PLANE   TRIANGLES.  t>5 

37.  From  the  top  of  a  hill  I  observe  two  milestones  in  the 
plain  below,  and  in  a  straight  line  before  me,  and  find  their  angles 
of  depression  5°  and  15°  ;  what  is  the  height  of  the1  hill? 

3<S.  Two  observers  on  the  same  side  of  a  balloon,  and  in  the 
same  vertical  plane  with  it,  are  a  mile  apart,  and  find  the  angles 
of  elevation  15°  and  G5°  30'  respectively ;  what  is  its  height? 

39.  Two  ships,  lying  half  a  mile  apart,  find  the  angles  sub- 
tended by  the  other  ship  and  a  fort,  respectively,  56°  19'  and 
G3°  14' ;  find  the  distance  of  each  ship  from  the  fort.     Prove  the 
work. 

40.  Bearings  at  sea  are  commonly  reckoned  in  points  and 
quarter-points,  thus : 

The  points  from  north  to  east  are :  x.,  x.  by  E.,  X.X.E.,  X.E. 
b}'  x.,  X.E.,  X.E.  b}T  E.,  E.X.E.,  E.  by  x.,  E.  ;  and  the  quarter-points 
are :  x.,  x.  J  E.,  x.  -J-  E.,  x.  f  E.,  x.  by  E.,  x.  by  E.  J  E.,  x.  b}' 

E.  -j-  E.,  x.  by  E.  f  E.,  X.X.E.  ;  X.E.  by  x.  f  x.,  ,  X.E.  ;  X.E.  £ 

E,  ,-  E.X.E.  ;  E.  by  x.  f  x.,  ,  E. 

Name  in  like  manner  the  points  and  quarter-points  from  x.  to 
w.  ;  from  s.  to  E.  ;  from  s.  to  w. 

How  many  degrees  in  four  points?  in  one  point?  in  one 
quarter-point?  * 

How  many  degrees  and  minutes  from  X.X.E.  to  E.  by  x.  ?  to 
E.  by  s.  ?  to  s.  by  E.  ?  to  s.  by  E.  f  E.  ? 

41.  A  privateer  lies  10  miles  s.w.  of  a  harbor,  and  observes 
a  merchantman  leave  it  in  the  direction  E.  by  s.,   at  the  rate 
of  9  miles  an  hour  ;  in  what  direction,  and  at  what  rate,  must  the 
privateer  sail  in  order  to  overtake  the  merchantman  in  1£  hours? 

42.  From  a  vessel  two  headlands  were  observed :    the  first 
bore  x.x.w.,  and  the  second  X.E.  by  E.  ;  then,  sailing  12  miles 
E.X.E.,  the  first  bore  x.w.  and  the  second  X.E.    Find  the  bearing 
and  distance  of  one  headland  from  the  other. 

43.  Find  the  ratio  of  the  areas  of  two  regular  decagons,  the 
one  inscribed  in,  and  the  other  circumscribed  about,  a  circle. 

44.  Find  the  angle  at  which  the  side  of  a  pjTamid  is  inclined 
to  the  base,  the  sides  being  equilateral  triangles  and  the  base  a 
square  ;  thence  find  the  diedral  angle  of  a  regular  octaedron. 


66  PLANE  TRIGONOMETRY.  [III. 

45.  Find  the  diedral  angle  of  an  edge,  the  perpendicular  from 
the  vertex  to  the  base,  and  the  distance  apart  of  two  opposite 
edges,  of  a  regular  tetraedron  whose  edge  is  unit}'. 

46.  If  ABC  be  an}'  plane  triangle,  express  sin  A,  cos  A,  tan  A, 
and  cos  A  -+-  COSB  cose,  in  terms  of  the  functions  of  B  and  c. 

47.  In  an}'  plane  triangle  ABC,  prove  that 

a  cos  B  +  b  cos  A  =  c  ; 
acosB  cose  4-  6  cose  cos  A  4-  c  cos  A  cos  B 
=  asinB  sine  =6sinc  sin  A  =  csinAsinB.  [Ex.46 

*48.  From  the  known  relations  of  the  parts  of  a  right  triangle 
GAP,  o+p=90°,  r2  =  arj+?/2,  a?  =  rcoso,  y  =  rsino,  prove  that 

x          y 
dp  =  —  do,        dr  =  -dx  +  -dy  =  coso  .  dx  +  sin  o  .  dy. 

dx  =  coso  .  dr  —  r  sin  o  .  do,     dy  =  sin  o  .  dr  -f-  rcos  o  .  do, 
and,  by  eliminating  dr  from  the  last  two  equations,  that 

sino  coso  —  ydx  4-  xdy 

do  =  ---  dx  -\  --  •  dy  =  —  '—s  -  9  —  '-•> 
r  r  x*  +  y2 

wherein  dx,  dy,  dr,  do,  dp  stand  for  total  differentials  of  x,  y, 
r,  o.  P  ;  i.e.,  for  any  simultaneous  infinitesimal  changes  in  the 
quantities  x,  y,  ^  o,  P,  that  are  consistent  with  the  known  rela- 
tions of  the  parts  of  a  right  triangle. 

*49.  If,  in  a  right  triangle,  only  the  values  of  x  and  y  be 
given,  and  if  these  have  the  possible  errors  ±  x'  and  ±  y1  respec- 
tively ;  i.e.,  if  x  may  possibly  differ  from  its  assumed  value  by 
either  4-  x1  or  —  x',  and  y  by  either  -f  y1  or  —  y',  these  signs 
not  being  necessarily  alike  ;  show  from  Ex.  48  that  the  resulting 
values  of  r  and  o  will  have  the  possible  errors 


and  ±  yx      xy  ,  =  ±~(x'  sino  +  yf  coso)  ; 

r2  r 

wherein  x'  and  y'  are  positive. 

So,  if  only  x  and  r  be  given,  with  the  possible  errors  ±  xf  and 
±r',  find  the  possible  errors  of  the  other  sides  and  angles. 

So,  if  only  x  and  o  be  given,  or  only  r  and  o,  with  the  pos- 
sible. errors  ±xf  and  ±of,  or  ±r'  and  ±o'. 


§  G.]  SOLUTION   OF   PL  AXE   TRIANGLES*.  67 

*50.    From  the  known  relations  of  the  parts  of  an  oblique 

triangle  ABC,  A  +  B  -f  c  =  180°,  ashiB  =  b  sin  A,  .....  ,  prove  that 

a]  dA-fdB-Mc  =  0, 

b~\  b  cos  A  .  dA.  —  a  cos  B  .  dv  —  sin  B  .  da  -f-  sin  A  .  db  =  0, 

c  COSB  .  cZB  —  b  cos  c  .  dc  —  sine  .  db  +  sinB  .  dc  =  0, 
a  cos  c  .  dc  —  c  cos  A  .  cLv  —  sin  A  .  do  4-  sine  .  da  =  0. 

From  these,  b}'  elimination  and  reduction  [104,  Ex.  46],  derive 

c]  b  .  dc  -f-  c  cos  A  .  dn  —  sin  A  .  dc  -f-  sin  c  .  da  =  0, 
c.dv  +  bcosA.dc  —  s'mA.db  -f-sinB.da  =  0, 

with  four  similar  equations,  which  maybe  written  from  S3rmmetry  ; 

d]  frsinc-.  dA  —  da  +  cose.  db  +  COSB.  dc  =  0,         [Ex.47 
with  two  similar  equations,  which  may  be  written  from  symmetry. 

*51.  If  in  an  oblique  triangle  only  side  a  and  angles  B,  c  be 
given,  and  if  their  possible  errors  be  ±  -  —  ,  ±  10",  and  ±15", 

respectively,  find-  the  possible  errors  of  A  [Ex.  50,  a]  ;  of  b 
[Ex.  50,  c]  ;  of  c  [Ex.  50,  c]. 

Find  the  values  of  these  possible  errors  when  ABC  is  very 
nearly  equilateral,  5000  feet  on  each  side. 

*52.  Given  the  values  of  c,  a,  6,  with  the  respective  possible 
errors  ±  cf,  ±  a',  ±  b\  deduce  the  possible  errors  of  B  and  A 
[Ex.  50,  c]  ;  of  c  [Ex.  50,  d]. 

*53.  Given  A,  a,  &,  with  possible  errors  ±  A',  ±a',  ±  &', 
deduce  the  possible  errors  of  B  [Ex.  50,  &]  ;  of  c  and  c. 

*54.  Given  A,  B,  6,  with  possible  errors  ±  A',  ±  B',  ±  6',  find 
the  possible  errors  of  c,  a  and  c  ;  first,  when,  as  in  all  the  above 
cases,  the  computation  is  assumed  to  be  exact  ;  and  secondly, 
when  c,  a,  c  are  liable  to  the  further  possible  errors  ±c",  ±a", 
±c"  from  omitted  decimal-figures,  etc.,  in  the  computation. 

*55.  Given  a,  6,  c,  with  possible  errors  ±a',  ±6',  ±c',  find  the 
possible  error  of  A  ;  the  possible  error  of  computation  being  A". 

NOTE.  If  ±a',  ±6',  ±  c',  ±  A"  denoted  "  probable  errors," 
the  probable  error  of  A  would  be 


In  the  same  way,  Ex.  49,  51-54  are  adapted  to  probable  errors. 


SPHERICAL   TRIGONOMETRY. 

IV.     SOLUTION   OF   SPHERICAL   TRIANGLES. 

§  1,      GEOMETEICAL    PEINCIPLES. 

If  about  the  vertex  of  a  triedral  angle  as  center  a  sphere  be 
described,  the  traces  [intersections]  of  the  three  faces  of  the 
triedral  upon  the  surface  of  the  sphere,  are  arcs  of  great  circles, 
and  together  the}'  constitute  a  spherical  triangle,  whose  sides 
measure  the  face-angles,  and  whose  angles  measure  the  diedral 
angles,  of  the  triedral.  SPHERICAL  TRIGONOMETRY  is  therefore 
the  trigonometry  of  the  triedral,  and  it  treats  of  the  numerical 
relations  of  the  six  parts  of  the  triedral,  viz.  :  the  three  face- 
angles  and  the  three  diedrals. 

As  the  three  faces  of  the  triedral,  when  produced,  are  in- 
definite planes,  and  divide  all  space  into  eight  solid  angles,  so 
the  sides  of  the  spherical  triangle,  when  produced,  are  great 
circles,  and  divide  the  surface  of  the  sphere  into  eight  triangles. 
Of  these  eight  triangles,  any  two  that  are  diametricall}*  opposite 
are  symmetrical,  and  anjr  two  not  opposite  have  one  side  and  its 
opposite  angle  the  same  in  each,  while  the  remaining  sides  and 
angles  of  the  one  are  supplementary  to  the  corresponding  sides 
and  angles  of  the  other. 

In  this  treatise  only  those  triangles  are  considered  whose  parts 
are  positive  and  each  less  than  180° ;  but  in  Astronomy  the 
general  spherical  triangle  is  used,  which  is  free  from  this  restric- 
tion. In  the  restricted  triangle  above  named  the  following 
principles  hold  true  ;  the}-  are  proved  in  Geometry,  and  grouped 
together  here  for  the  convenience  of  the  reader : 

The  sum  of  the  three  sides  lies  between  0°  and  360°. 

The  sum  of  the  three  angles  lies  between  180°  and  540°. 

Each  side  is  less  than  the  sum  of  the  other  two. 


§  1.]  SOLUTION   OF   SPHERICAL   Til  TANGLES.  69 

Each  angle  is  greater  than  the  difference  between  180°  and  the 
sum  of  the  other  two. 

Of  any  two  unequal  sides,  the  greater  lies  opposite  the  greater 
angle ;  and  conversely. 

Each  side  or  angle  is  the  supplement  of  the  corresponding 
angle  or  side  of  the  polar  triangle. 

If  two  sides  of  a  triangle  are  equal,  so  also  are  the  opposite 
angles;  and  conversely. 

The  perpendicular  from  the  vertex  of  an  isosceles  triangle  to 
the  base  bisects  both  the  vertical  angle  and  the  base. 

Equilateral  triangles,  in  general,  are  not  similar. 

A  spherical  triangle  is,  in  general,  determined  when  any  three 
of  its  six  parts  are  known. 

The  area  is  to  the  hemisphere  as  the  excess  of  the  sum  of  the 
angles  over  180°  is  to  360°. 

The  following  principles,  proved  later,  are  added : 

In  a  right  spherical  triangle  :  An  oblique  angle  and  its  opposite 
side  are  of  the  same  species  [both  less  than  90°,  both  greater 
than  90°,  or  both  equal  to  90°].  [Thm.  1  Cor.  1 

If  the  Irypothenuse  is  less  than  90°,  the  other  two  sides  are  of 
the  same  species,  and  so  are  the  two  oblique  angles  ;  but  if  the 
liypothenuse  is  greater  than  90°  they  are  of  different  species,  and 
so  are  the  two  oblique  angles.  [Thm.  1  Cor.  2 

If  the  hypotheuuse  equals  90°,  another  side  and  its  opposite 
angle  are  each  equal  to  90°  ;  and  if  another  side  or  its  opposite 
angle  equals  90°,  the  hypothenuse  equals  90°.  [Thm.  1  Cor.  3 

No  side  is  nearer  to  90°  than  its  opposite  angle.   [Thm.  1  Cor.  \- 

In  an}'  spherical  triangle  :  A  side  that  differs  from  90°  not 
less  than  another  side  is  of  the  same  species  as  its  opposite 
angle.  [Thm.  3. Cor. 

An  angle  that  differs  from  90°  not  less  than  another  angle  is 
of  the  same  species  as  its  opposite  side.  [Thm.  4  Cor.  1 

There  are  at  least  two  sides  which  are  of  the  same  species  as 
their  opposite  angles.  [Thin.  4  Cor.  2 

The  half-sum  of  any  two  sides  and  the  half-sum  of  their  oppo- 
site angles  are  of  the  same  species.  [Thm.  12  Cor. 


70 


SPHERICAL   TRIGONOMETRY. 


[IV. 


§  2,     NAPIEE'S  EULES  FOE  THE  EIGHT  TEIANGLE. 

THEOREM  1.  In  a  right  spherical  triangle,  if  the  right  angle 
be  ignored,  and  if,  of  the  five  remaining  parts,  the  two  sides  be 
taken,  and  for  the  hypothenuse  and  the  two  oblique  angles  their 
complements  be  substituted,  then  the  sine  of  any  one  of  the  five 
parts  (called  the  middle  part)  equals  :  ' 

(1)  The  product  of  the  cosines  of  the  two  opposite  parts; 

(2)  The  product  of  the  tangents  of  the  two  adjacent  parts. 

[The  figure  shows  four 
right  spherical  triangles  : 

PQR, 
p,  q  and  r  all  <  90°; 

PQR', 
j>'andg'>90°,  r<90°; 

P'QR, 


2>'andrf>90°,  g'"<90°. 
The  demonstration  applies 
to  all  alike.] 

Let  PQR  be  anjT  spherical  triangle,  wherein  R  is^a  right  angle  ; 
p  and  Q  oblique  angles,  either  acute  or  obtuse  ;  r  the  hypothe- 
nuse ;  p  and  q  sides  opposite  P  and  Q  respectivel}'  ;  then  will  : 


151] 

15-2] 
153] 
154] 
155] 


sin  p  =  sin  r  sin  p  and 

sin  q  —  sin  r  sin  Q  and 

cos  r  =  cosp  cos  q  and 

cos  P  =  cosp  sin  Q  and 

cos  Q  =  cos  q  sin  p  and 


=  tang  cotQ ; 
=  tanp  cotp ; 
=  cotp  cotQ; 
=  tan  q  cot  r ; 
=  tanpcotr. 


For,  join  the  vertices  p,  Q,  R  to  o,  the  center  of  the  sphere, 
and  through  either  oblique  angle,  as  P,  draw  PA  and  PB,  J_  OP, 
and  meeting  OR  and  OQ  in  A  and  B  ; 

then  also  is  AB_LOA  and  PA,  \_geom. 

and  A  OPA,  OPB,  OAB  and  PAB  are  right-angled  at  P,  P,  A  and  A. 


§  2.]  SOLUTION   OF   SPHERICAL   TRIANGLES.  71 

But  arcs  p,  q  and  r  measure  Z^AOB,  POA  and  FOB, 

[geom. 


.-.   sinp  =  —  ,  cosp  =  —  ,  tanp  =  —  ; 

OB  OB  OA 

T>  A  OT*  PA. 

sing  =-,  cosg  =  -,  tang  =  -; 


sn,-   =      , 

OB  .OB                            OP 

sinp   =^,  COSP  =  ^,  tanp  =  ^.         -        [I.  §  7 

PB  PB                             PA 


OB          OB     PB 

sing,  =  —  ,  =  —  .~  =  tanpcotp; 

OA          OA     AB 

cosr,  =^,  =  ^.2Z  = 

OB          OB     OA 

COSP,  =—,  =  —  .  —  =  tangcotr.  Q.E.D. 

PB          OP     PB 

So,    •••   either  oblique  angle  may  be  p  and  the  other  Q, 
.•.    sin  q  =  sinr  sinQ  ; 
sin  p  =  tan  q  cot  Q; 
COSQ  =  tanp  cot?*.  Q.  E.  D. 


And,-.'   cotp  =       2-     and    cotQ  =         ,  [above 

tanj)  tang 


.-.   cotp  cotQ  =  sin^  sn  q  =  cosp  cosg  ;  [35 
tanp  tan  g 

but           cosr  =  cos^>  cosg,  [above 

.-.    cosr  =  cotp  cotg.  Q-  E.  D 


And,  •.•    COSP  =  tang  cot  r,    COSJP  =       -,    sin  Q  =         ,  [above 

cosg  smr 

.  •  .    cos  P  =  cosp  sin  Q. 
So,  cos  Q  =  cos  g  sin  p.  Q.  E.  D. 

COR.  1.  In  a  right  spherical  triangle,  an  oblique  angle  and  its 
opposite  side  are  of  the  same  species  [both  less  than  90°,  both 
greater  than  90°,  or  both  equal  to  90°]. 


72  SPHEKICAL   TIUGOXOMETKY.  [IV. 

For,  v    cos  p  =  cosp  sin  Q,  [154 

and    v    shiQ  is  not  0,  and  is  always  positive,      [§  1, 1.  Thm.3 

.'.    COSP  and  cosp  are  both  positive,  or  both  negative,  or 
both  zero ; 

.'.    P  andp  are  both  <  90°,  or  both  >  90°,  or  both  =  90°. 
So,  Q  and  q  are Q.  E.  D. 

COR.  2.  In  a  right  spherical  triangle,  if  the  liypothenuse  is 
less  than  90°,  the  other  two  sides  are  of  the  same  species,  and  so 
are  the  two  oblique  angles;  but,  if  the  liypothenuse  is  greater  than 
90°,  they  are  of  different  species,  and  so  are  the  two  oblique  angles. 
For,  v  all  parts  of  a  triangle  are  positive  and  all  less  than  180°, 
and  v  cos  r  =  cosp  cos  q,  and  =  cotpcotQ,  [153 

.*.    if  cos?*  is  positive,  then  cosp  and  cos  q  are  both  posi- 
tive or  both  negative  ;  and  so  are  cotp  and  cotQ  ; 
i.e.,          if  r  <  90°,  p  and  q  are  both  <  90°,  or  both  >  90°,  and  so 
are  p  and  Q  ;  Q.  E.  D.    [I.  Thm.  3 

but,          if  cosr  is  negative,  then  cosp  and  cosg  are  one  positive 

and  the  other  negative  ;  and  so  are  cot  p  and  cot  Q  ; 
i.e.,          if  r>90°,  p  and  q  are  one  of  them  <  90°  and  the  other 
>  90°,  and  so  are  p  and  Q.  Q.  E.  D.  [I.  Thm.  3 

COR.  3.    In  a  right  spherical  triangle,  if  the  hypothenuse  equals 
90°,  another  side  and  its  opposite  angle  are  also  each  equal  to  90°; 
and  conversely,  if  another  side  or  its  opposite  angle  equals  90°, 
the  hypothenuse  equals  90°. 
For,         ifcosr=0,  then  cosp  cos  <?  =  0,  and  cotp  cot Q  =  0,  [153 

.'.    p  or  g=90°,     and    p  or  Q  =90°,  [I.  Thm.  14 

i.e.,          whichever  of  the  sides  and  angles  p,  q,  p  or  Q  =  90°,  the 
opposite  angle  or  side  =  90°.  .  Q.  E.  D.    [Cor.  1 

So,  ifporp=90°,    then    cosp  =  0    or    cotp  =  0, 

/.    cosr  =  0,     and     r  —  90°.  [153 

So,  ifgorQ  =  90°,      r=90°.  Q.E.D. 

COR.  4.    In  a  right  spherical  triangle,  no  side  is  nearer  to  90° 
than  its  opposite  angle. 
For,         if  p~90°<p~90°, 
then         sing,  =tanpcotp,  =tanp  :  taiip,  >1;  which  is  absurd. 


§3.] 


SOLUTION   OF   SPHERICAL  TRIANGLES. 


73 


§  3,      GENERAL    PROPERTIES   OF  SPHERICAL  TRIANGLES. 

TIIM.  2.    In  any  spherical  triangle  the  sines  of  the  sides  are 
proportional  to  the  sines  of  the  opposite  angles. 


\c 


[151 


•  D 


Let  ABC  be  any  spherical  triangle  ;  then  will : 
156]         sina:  sin&  =  sinA:  sinB  ; 
sin  b  :  sin  c  =  sin  B  :  sin  c  ; 
sin  c  :  sin  a  =  sin  c  :  sin  A. 
For,  draw  CD,  =  |>,  _L  AB  ;  then  : 

sinp  =  sinasinB,     and     sinjp  =  sin&shiA, 
sin  a  sin  B   =  sin  b  sin  A. 

.-.     sin  a  :  sin  b  =  sin  A  :  sinB.  [Thm.  prop'n 

sin  b  :  sin  c  =  sin  B  :  sine  ; 
sine  :  sin  a  =  sine  :  sin  A.  Q.  E.  D. 

NOTE.    This  theorem  may  be  stated  more  symmetrically  thus  : 

sina  :  sin6  :  sine  =  sinA  :  sinB  :  sine  ; 
or  thus : 

sina  _  sin  b  __  sine 
sinB 


So, 
and 


sin  A 


sine 


COR.    In  any  spherical  triangle  ABC  : 


sin  a  = 


sin  b  sin  A      sine  sin  A 


smA  = 


sin  B  sin  a      sm  c  sin  a 


sin  b  = 


sin  c  = 


S111B 

sincsiiiB 

sine 
sin  a  sin  c 

sin  A 


sine 
sin  a  sin  B 

sin  A 
sin  &  sine 

sinB 


S111B  = 


smc  = 


sin  b 
sin  c  sin  b 

sine 
sin  A  sin  c 

sina 


sine 
sin  A  sin  b 

sina 
sin  B  sin  c 

sin  b 


74 


SPHERICAL   TBIGQ^OMETBY. 


[IV. 


TIJM.  3.    In  any  spherical  triangle  ABC  : 
157]         cos  a  =  cos  b  cose  -f-  sin  b  sine  cos  A, 
cos  6  =  cos  c  cos  a  -f-  sin  c  sin  a  cos  B, 
cos  c  =  cos  a  cos  &  -f-  sin  a  sin  &  cos  c. 


[The  figure  shows  four  oblique  spherical  triangles : 
ABC,  a,  b  and  c  all  <  90° ;  AB'C,  a'  and  c'  >  90°,  b  <  90°  ; 

A'BC,  a  <  90°,  6'  and  c"  >  90°  ;     A'B'C,  a'  and  6'  >  90°,  c'"  <  90°. 

The  demonstration  applies  to  all  alike.] 

For,  join  the  vertices  ABC  to  o,  the  center  of  the  sphere,  and 

through  A  draw  AE  and  AF,  J_OA,  and  meeting  OB  and  oc  in 

E  and  F  ;  then  A  OAE  and  OAF  are  right-angled  at  A. 

But  arcs  a,  b  and  c  measure  A  EOF,  AOF  and  AOE, 

and  A  =  /.  EAF  ;  then  : 


cosa  = 


OE2  +  OF2  —  EFS 


and 


2  OE  .  OF 
OE2  =  OA2  -f  AE2,     and    OF2  =  OA2  -f-  AF2  ; 

OA2  4-  AE2  +  OA2  +  AF2  —  EF2 
2  OE  .  OF 


[geom. 
[108 
\_geom. 


cosa  = 


OA* 
OE  .OF 


AE 


2+AF2_EF2 
2  OE  .  OF 


AF2  —  EF2 


_  OA     OA    ,    AF     AE      AE 

OF      OE        OF      OE  2  AE  .  AF 


But 


COSA  = 


AE2  +  AF2  -  EF2 
2  AE  .  AF 


[108 


§3.] 


SOLUTION   OF   SPHERICAL   TRIANGLES. 


75 


.  • .     cos  a  =  cos  b  cos  c  -f-  sm  &  sin  c  cos  A. 
So,  cos  b  ==  cose  cos  a  -f-  sine  sin  a  cos  B, 

and          cos  c  =  cos  a  cos  6  +  sin  a  sin  6  cos  c. 


[I.  §7 

Q.E.  D. 


COR.   In  any  spherical  triangle,  a  side  that  differs  from  90°  not 
less  than  another  side,  is  of  the  same  species  as  its  opposite  angle. 


-r,  COS « —  COS  0  COSC  r-irrrr 

lor,  •.*     COSA  =  —  — »  1157 

sm  b  sin  c 

and  •.•  the  product  sin  b  sine  is  always  positive,        [I.  Thm.  3 

.  *.  cos  A  has  the  same  sign  as  (cos  a  —  cos  b  cos  c) . 

But  if  a  differs  from  90°  not  less  than  6  or  c, 

then  cos  a  is  as  large  [great  numerically]  as  cos  b  or  cose, 

and  if  cos  a  =  0,     then     cos  b  or  cos  c  =  0  ; 

and  • .  *  cos  b  and  cos  c  are  both  smaller  [less  numerically]  than  1, 

.-.  cos  a  is  larger  than  cosb  cose  ; 

.• .  cos  a  gives  sign  to  (cos  a  —  cos  b  cos  e) ; 

.*.  cos  A  has  the  same  sign  as  cos  a,  [above 

or  ifcosa  =  0,     then     COSA  =  O; 

i.e.,  both  are  positive,  or  both  negative,  or  both  zero ; 

.-.  A  and  a,  both  <  90°,  or  both  >  90°,  or  both  =90°.  Q.  E.  D. 

THM.  4.    In  any  spherical  tri- 
angle ABC  : 

158]         cos  A  =  —  cos  B  cos  c 

H-sinB  sine  cos  a, 

COS  B  =  —  COS  C  COS  A 

-|-  sine  sin  A  cos&, 

COS  C  =  —  COS  A  COS  B 

-j-sinA  siiiB  cose. 
For,  let  the  triangle  A'B'C'  be 
polar  to  ABC  ; 

then,-.-     a'=7r— A,  &'=TT— B,  C'=TT— c,   and  A'=TT— a,  [geom* 
and  •••     cosa'=  cos  6' cos  c'  + sin  6' sine' cos  A',  [157 

.-.  —  cos  A  =  (—  COSB)  (—cose)  -f-  sin  B  sinc(— cosa),  [10,11 
.-.     COSA  =  —  COSB  cose +  sinB  sine  cosa. 
So,  cos  B  =  —  cos  c  cos  A  +  sin  c  sin  A  cos  &, 

and  cose  =  —  COSA  COSB  -f-  sin  A  sin  B  cose.  Q.E.  D. 


76  SPHERICAL  TRIGONOMETRY.  [IV. 

COR.  1.  In  any  spherical  triangle,  an  angle  that  differs  from 
90°  not  less  than  another  angle,  is  of  the  same  species  as  its  oppo- 
site side. 


For,  ...     cos«=  .  [158 

sm  B  sin  c 

and  v  the  product  sin  B  sine  is  alwa}rs  positive,         [I.  Tbm.3 

cos  a  has  the  same  sign  as  (cos  A  +  COSB  cose). 

But  if  A  differs  from  90°  not  less  than  B  or  c, 

then  cos  A  is  at  least  as  large  as  COSB  or  cose, 

and  if  cos  A  =  0,     then     COSB  or  cose  =  0  ; 

and  •.•  COSB  and  cose  are  both  smaller  than  1, 

.  •  .  cos  A  is  larger  than  cos  B  cos  c  ; 

.-.  cos  A  gives  sign  to  (cos  A  -f-  COSB  cose)  ; 

.•.  cos  a  has  the  same  sign  as  cos  A,  [above 

or  if  cos  A  =  0     then     cos  a  =  0  ; 

i.e.,  both  are  positive,  or  both  negative,  or  both  zero  ; 

.-.  a  and  A  are  both  <  90°,  or  both  >  90°,  or  both  =  90°. 

Q.E.D.    [I.  Thms.3,14 

COR.  2.    In  any  spherical  triangle  there  are  at  least  two  sides 
which  are  of  the  same  species  as  their  opposite  angles. 

This  is  a  direct  consequence  of  Thm.  3  Cor.  and  Thm.  4  Cor.  1. 
THM.  5.    In  any  spherical  triangle  ABC  : 
)sin(s  —  6)  sin  (8  —  c) 


sn    B 


For,  •.•     2  sin2|  A  =  1— cos  A  [63 

_  .. cos  a  —  cos  b  cos  c 

sin  b  sine 
_  cos  b  cos  c  +  sin  b  sine  —  cos  a 

sin  b  sine 

__  cos  (b  —  c)  —  cos  a 
sin  b  sine 


I  sin  (s  —  c),sin(s  —  a) 
=  \  -      —  .        —      —  •  » 
\  sine  sin  a 

sin  (s  —  a)  sin(s  —  b)        r       *,    .,  .    , 
-    ' 


[42 


3.]  SOLUTION   OF   SPHERICAL   TRIANGLES.  77 

—  2  sin  1(6  —  c  +  fl)sini(6  —  c  —  a)        ,-71 

sin  b  sine 
2  sin  i  (a  — 


So,  sin 


sin  b  sine 
2  sin  (s  —  b)  sin  (s  —  c) 

sin  b  sine 
sin  (s  —  6)  sin  (s  —  c) 

s'mb  sine 

sin  (s  —  c)  sin  (s  —  a) 
sine  sin  a 


1  sin  (s  —  a)  sin  (s  —  b) 
=  V~     '    sinasinft 

THM.  6.     In  any  spherical  triangle  ABC  : 

Isin  s  sin  (s  —  a) 

1601         cosi-A     =^l i -i 

\      sin  0  sine 

[sin  s  sin  (s  —  6) 

COS4-B        =  .* -j 

\       sine  sin  a 

cos^c      as.j__ L 

\       sin  a  si 


For,*/     2cos2^A=  1  4- COSA  [64 

,  cos  a—  cos  6  cose 


sin  b  sin  c 
cos  a  —  cos  b  cos  c  -f-  sin  6  sin  c 

sin  b  sin  c 
COSCT  —  cos(6-fc) 
sin  b  sin  c 


sin 6  sine 
'a  4. 5  -|-  c)  sin|(  — a  +  5  +  c) 

sin  6  sin  c 
2  sins  sin  (s  —  a) 

sin  b  sin  c 
sin  s  sin  (s— a) 
sin  b  sine 


ri --7 
L1O/ 


[41 


78  SPHERICAL    TRIGOXOMETKY.  [IV. 

(sin .9  sin (5  —b) 
So,  COS^B  =xv~ 


sin  c  sin  a 


(sins  sin(s  —  c) 
=  \     siuasiuft 

THM.  7.     JTI  cmy  spherical  triangle  ABC  : 

161]         tan|A  =  ./si"(s-6)sin^-c), 

\      sin  s  sin  (s  —  a) 

-     Isin  (g~c)  sin  (8-g) 

\      sins  sin(s—  6) 


. 

m  (s  —  c) 

The  reader  may  prove  the  theorem  by  dividing  the  value  of 
sin  IA  by  that  of  COS|-A  [Thins.  5,  6]  ;  and  so  for  tan  IB  and 
tan^c. 

THM.  8.    In  any  spherical  triangle  ABC  : 

I—  cos  s  cos  (s  —A) 
162]         sin|a  =  -J- 


sin  B  sine 

—  cos  s  cos  (s  —  B) 
sine  sin  A 


—  cos  s  cos  (s  —  c)  r       ., 

- 


For,  let  the  triangle  A'B'C'  be  polar  to  ABC  : 

then  v     a'  =  7T  —  A,     b'  —  7r  —  B,     C'  =  TT  —  c, 

>•  f  aeom. 

A'  =  7T—  a,         B'  =  7T  —  0,         C'  =  7T  —  C, 

/sins'  sin  (sf  —  a') 
and  -.. 


sin  6' sine' 


— cos  s  cos  (s  — A)  _ 

V  [0,16,4,10 


3.J  SOLUTION  OF   SPHERICAL   TRIANGLES.  79 

I—  cos  s  cos  (s  —  B) 


So,  sin  &  b  = 


sine  sn  A 


I—  coss  cos(s  —  c) 
sin^c  =  V sinAsinB  <*•*•»• 

The  reader  may  also  prove  the  theorem  directly  from  the 
formulae  of  Thm.  4  : 

cos  A  =  —  cos  B  cos  c  +  sin  B  sin  c  cos  a 

in  the  same  manner  as  Thm.  5  was  proved  from  the  formulae  of 
Thm.  3. 

THM.  9.    In  any  spherical  triangle  ABC  : 
cos  (s  —  B)  cos  (s  —  c) 


1631 

sin  B  sine 

Icos  (s  —  c)  cos  (s  —  A) 


I 
=  \ 


sne  sn  A 


cos  (s  —  A)  cos  (s  —  B)         r        i  ,  >. 

-        siulsmB 

The  reader  may  prove  the  theorem  directly  from  the  formulae 
in  Thm.  4,  or  prove  it  by  aid  of  the  polar  triangle  in  the  same 
manner  as  Thm.  8  was  proved. 

THM.  .10.     In  any  spherical  triangle  ABC  : 


164]         taniq  =  J 

\cos(s—  B)  cos(s—  c) 

=J  -cosscoS(s-B) 
\cos(s—  c)cos  (s—  A) 
ta  J    -cosscos(s-c) 

\cos(s  —  A)COS(S  —  B) 

The  reader  may  prove  the  theorem  by  dividing  the  value  of 
sin^-a  by  that  of  cos^a  [Thms.  8,  9]  ;  and  so  for  tan  ^6  and 
tan  -J  c. 

NOTE.  The  radicals  in  [159-164]  are  all  positive  ;  for  they 
are  the  sines,  cosines  and  tangents  of  angles  £A,  .....  Ja,  .....  , 
which  are  all  in  the  first  quadrant. 


80  SPHEEICAL   THIGOXOMETKY. 

THM.  11.     In  any  spherical  triangle  ABC  : 
165]         sinKA  +  B)  =  ^ig^)coS}C; 

1GC]         sini(A~B) 
1C7]         CosKA  +  B) 


V. 


. 

sinc 


cos-i-c 


1G8]         coSKA~B)  =  ,  . 

sm-|c 

For,  v     sin  £(A  ±  B)  =  sin(iA  ±  JB) 

=  sin-i-Acos-i-B  ±  cos|AsiniB        [39,40 

_siii(s—  b)    |sinssin(g—  c)     sin(g—  a)    /sinssin(s—  c)  pj.^  1GQ 
sine      \     sin  a  sin  b  sine      \     sin  a  sin  6 

sin(s  —  b)  ±sin(s—  a) 
—          ' 


sine 


2cos4-csini(a  —  6) 


zsin^c  cos^-c 


and 


i 
sm-J-c 


[160 
[68,  60 
[69,60 

Q.E.D. 

[41,42 


So,   •.•     COS|-(A  ±B)  =  co 
_sins    !sin(s—  a)  sin(s—  b)      sin  (s—  c)    Jsin  (s—  a)  sin  (s—  b} 
sinc\          sin  a  sin  b  sine       \  sin  a  sin  b 

sin  (s-c)sinic  [159 


sns 


sine 


and 


cos^c 


sin  Tr 


. 
sm-Jc, 

.    .. 
sm|c. 


Q.E.D. 


§3.]  SOLUTION   OF   SPHERICAL   TRIANGLES.  81 

These  four  formulae,  with  the  like  formulae  found  when  the  other 
sides  and  angles  are  employed,  are  called  Delambre's  Formulae. 

THM.  12.    In  any  spherical  triangle  ABC  : 


171] 

172] 

The  reader  ma}'  prove  the  theorem  by  dividing  the  formulae 
of  Thm.  1  1  one  by  another,  viz.  : 

[165]  by  [167]  ;  [166]  by  [168]  ; 

[168]  by  [167]  ;  [166]  by  [165]. 

These  four  formulae,  with  the  like  formulae  found  when  the 
other  sides  and  angles  are  employed,  are  called  Napier'  s  Analogies. 
COR.    In  any  spherical  triangle  the  sum  of  any  two  sides  is 
less  than,  equal  to,  or  greater  than,  180°,  according  as  the  sum  of 
the  opposite  angles  is  less  than,  equal  to,  or  greatef~than,  180°. 
For,  v     tan  £  (a  +  6)  COS^(A  +  B)  =  tan^-c  COS£(A~B),     [171 
and   v     £c  and  £  (A  ~  B)  are  both  less  than  90°,  and  ^c  is  not  0, 
whence    the  product  tan^-c  COS£(A~B)  is  positive,  and  not  0  ; 
.'.     the  product  tan|-(a  -+-  b)  cos  ^(A  4-  B)  is  positive,  not  0, 
.*.     tan  J(a  +  &)  and  cos  J(A  +  B)  are  both  positive  or  both 

negative,  or  one  is  GO  and  the  other  0  ; 
.'.     £(a  +  6)  and  i(A  +  B)  are  both  <  90°,  both  =  90°,  or 
both  >  90°;  [I.  Thin.  3 

.'.     (a-f  6)  and  (A  +  B)  are  both  <  180°,  both  =  180°,  or 
both  >  180°.  Q.E.D. 

NOTE.  Thm.  1  Cor.  1,  and  the  Cors.  to  Thins.  3,  4  arid  12 
are  summarized  as  follows  : 

The  half-sum  of  any  two  sides  and  the  half-sum  of  their 
opposite  angles  are  of  the  same  species  ;  and  so  are  either  side 
and  its  opposite  angle,  unless  the  side  be  nearer  to  90°  than  an}* 
other  side,  and  the  angle  be  nearer  to  90°  than  an}-  other  angle. 


82  SPHERICAL  TRIGONOMETRY.  [IV. 

§  4,     SOLUTION   OF   EIGHT   TRIANGLES. 

PROB.  1.     To  SOLVE  A  RIGHT  SPHERICAL  TRIANGLE. 

Let  PQR  be  any  spherical  triangle  right-angled  at  R  ;  then  the 
formulae  of  Thm.  1  (Napier's  rules)  apply  directly,  and  with 
the  right  angle  and  an}'  other  two  parts  given,  the 
remaining  parts  can  be  found.  The  computer  will 
form  equations  containing  three  parts,  two  known 
and  one  unknown,  and  then  solve  these  equations 
for  the  unknown  part.  He  will  observe  that : 
If  the  three  parts  are  contiguous  to  each  other, 
then 

that  which  lies  between  the  other  two  is  the  middle 
part,  and  the  others  are  adjacent  parts  ; 
but  if  two  parts  lie  together  and  the  other  apart 
from  them,  then 

that  which  lies  apart  from  the  other  two  is  the  middle  part,  and 
the  others  are  opposite  parts. 

The  following  rule  will  solve  all  cases  : 

Take  each  of  the  two  given  parts  in  turn  for  middle  part,  and 
apply  that  one  of  Napier's  rules  which  brings  in  the  other  given 
part. 

Take  the  remaining  part  for  middle  part,  and  apply  that  one 
of  Napier's  rules  which  brings  in  both  of  the  parts  just  found. 

For  a  check  make  the  part  last  found  the  middle  part,  and 
apply  that  one  of  Napier's  rules  wJiich  brings  in  both  the  given 
parts. 

The  whole  work  requires  but  nine  logarithms,  or  seven  without 
the  check,  since  two  of  the  logarithmic  functions  are  used  twice 
over.  The  check  is  to  be  applied  to  the  sine  of  the  part  last 
found.  If  the  two  values  got  for  this  sine,  natural  or  logar- 
ithmic, differ  ~by  not  more  than  three  units  in  the  last  decimal 
place,  the  work  is  probably  right,  since  the  defects  of  the  tables 
permit  this  discrepancy  in  the  two  results.  If  such  discrepancy 
exist  the  mean  of  the  two  values  may  be  used. 


§  4.]  SOLUTION  OP   SPHElilCAL   TRIANGLES.  83 

CASE  1.    Given  p  and  q,  the  two  sides  about  the  right  angle, 

then: 

sing  =  tanpcotp,     ./.     cot  p  =  cotp  sing  ; 

sinp  =  tang  cot  Q,      .'.     cotQ  =  sinp  cot  q  ; 
cos  r  =  cot  P  cot  Q  ;  check  cos  r  =  cosp  cos  g. 
NOTE.   One  triangle  is  alwa}~s  possible,  and  but  one  ;  the  parts 
q,  P  and  Q  are  determined  without  ambiguity  by  the  formulae. 

CASE  2.     Given  the  hypothenuse  and  one  side,  for  example  p, 

then: 

cosr  =  cosp  cosg,      .'.     cos#  =  seep* cos r\ 

s'mp  =  sin  r  sin  p,        /.     sin  p  =  s'mp  esc  r ; 
cos  Q  =  cos  q  sin  p  ;   check  cos  Q  =  tanp  cot  r . 

NOTE.    A  triangle  is  possible  only  when  r  is  nearer  to  90° 
than  p  is,  or  when  r  andp  are  both  90°  ; 
for  then  only  can 

cosr  tan/? 

cos  q,  = ,<1;     or    COSQ,  =——,<!. 

cosp  tan  r 

The  formula  would  give  two  values  to  P,  an  acute  angle  and 
an  obtuse  one,  supplementary  to  each  other  ; 
but,  since  p  and  P  are  both  of  the  same  species,  £Thm.  1  Cor.  1 
onty  one  value  of  p  is  admissible.  The  parts  <?,  p  and  Q  are 
therefore  determined  without  ambiguity  by  the  formulae,  unless 
/>  and  r  are  both  90°,  when  COSQ  and  cos<?  become  indeterminate, 
and  q  =  Q  and  P  =  90°. 

CASE  3.    Given  an  oblique  angle  and  the  adjacent  side,  for 
example  p  and  q,  then : 

sin  q  =  tanjp  cotp,      .*.     taup  =  sin  q  tanp  ; 

cos  P  =  tan  q  cot  r,       .'.     cot?"  =  cot  q  COSP  ; 

cos  Q  =  tanp  cot  r  ;  check  cos  Q  =  cos  q  sin  P. 
NOTE.    One  triangle  is  alwa}'s  possible,  and  but  one  ;  the  parts 
p,  r  and  Q  arc  determined  without  ambiguit}'  by  the  formulae. 
•     CASE  4.     Given  an  oblique  angle  and  the  opposite  side,  for 
example  p  and  p,  then  : 

sinp  =  sin  r  ship,      .*.     sinr  =  sin_p  cscr  ; 

cos  P  =  cosp  sin  Q  ,     .  * .     sin  Q  =  seep  cos  p  ; 

sin  q  =  sin  r  sin  Q  ;  check  sin  q  =  tanp  cot  p. 


84  SPHERICAL   TRIGONOMETRY.  [IV. 

NOTE.    If  p  and  P  are  not  of  the  same  species,  no  triangle  is 
possible,  [Thin.  1  Cor.  1 

nor  if  p  is  nearer  90°  than  p.  [Thm.  1  Cor.  4 

If         p  arid  P  are  equal,  but  not  90°, 
then         sing,  sinr,  and  sinQ,  all,  =  1  ; 

.-.     </,  r,  and  Q,  all,  =  90°,  [I.  Thin.  14 

and  the  triangle  is  biquadrantal. 

If         p  and  P  are  both  90°, 

then         r  is  also  90°,  and  the  triangle  is  biquadrantal,     [geom. 
and          q  and  Q  are  indeterminate. 

If         P  is  nearer  90°  than  p, 
then  two  triangles  are  always  possible, 
for     v     g,  r  and  Q  are  all  determined  from  their  sines, 
and   v     to  every  sine  correspond  two  angles,  supplements  of 
each  other,  [10 

.'.     g,  r  and  Q  may  each  have  two  values  ; 

but,  v     q  and  Q  are  of  the  same  species,  [Thin.  1  Cor.  1 

.'.  if  r  is  less  than  90°,  q  and  Q  are  of  the  same  species 
with  p  and  P,  and  there  is  but  one  value  for  each  of 
them.  [Thin.  1  Cor.  2 

So,       if  r  is  greater  than  90°,  q  and  Q  are  of  the  opposite  spe- 
cies to  p  and  P,  and  there  is  but  one  value  for  each 
j    of  them ; 
.*.    two  triangles,  and  but  two,  are  formed  with  the  given 

data,  viz. : 

(1)  that  wherein  r  <  90°,  and  q  and  Q  are  of  the  same  species 
with  p  and  P. 

P  0  p,       (2)  that  wherein  r  >  90°, 

and  q  and  Q  are  of  the  op- 
posite species  to  p  and  P. 

This  ambiguity   appears 
directly  from  the  figure. 

For,  let  PQR  be  a  spheri- 
cal triangle  right-angled  at 
R  ;  produce  the  arcs  PQ  and  PR  to  meet  at  pr ;  then : 


§  4.]  SOLUTION  OF   SPHERICAL   TRIANGLES.  85 

and  •.•     A  PRQ  and  P'RQ  are  both  right  angles,  [hypoth. 

.*.     two.  right  triangles  exist,  PQU  and  P'QR,  which  have  the 

same  two  parts  given,  p  and  p,  and  the  remaining 

parts  of  the  one  triangle  supplementary  to  those  of 

the  other. 

CASE  5.    Given  the  hypothenuse  and  one  oblique  angle,  for  ex- 
ample P,  then  : 

cos  P  =  tang  cot  r,  .'.  tan  q  =  tan  r  cos  p  ; 
cosr  =  cotPcotQ,  .'.  cot  Q  =  cosr  tan  P; 
=  tang  cot  Q  ;  check  smp  ==.  sin  r  sin  p. 


NOTE.  One  triangle  is  always  possible,  and  but  one.  The 
part  p  is  of  the  same  species  with  p,  and  so  can  have  but  one  of 
two  possible  values  ;  the  parts  q  and  Q  are  determined  without 
ambiguity  by  the  formulae. 

CASE  6.    Given  the  two  oblique  angles  p  and  Q,  then  : 
cos  P  =  cosjp  sin  Q,      .  •  .    cosp  =  cos  p  esc  Q  ; 
cos  Q  =  cos  g  ship,      .*.    cosg  =  CSCPCOSQ  ; 
cos  r  =  cosp  cos  q  ;  check  cos  r  =  cot  P  cot  Q. 

NOTE  1.     The  parts  p,  q  and  r  are  determined  without  arnbi- 
guit}'  by  the  formulae  ;  but  the  solution  is  possible  only  when 
COSP  CSCQ,  CSCP  COSQ  and  cotPcotQ  are  each  smaller  than  1  ; 
i.e.,  when  COSP  is  smaller  than  sing  and  COSQ  is  smaller 

than  siup, 

and  this  when  p  is  nearer  to  90°  than  Q  to  0°  or  180°, 
and          when  Q  is  nearer  to  90°  than  p  to  0°  or  180°. 

NOTE  2.  The  computer  may  follow  a  different  order  from  that 
given  ;  he  ma}',  at  pleasure,  find  all  the  required  parts  directly 
from  the  given  parts,  or  compute  an}r  one  of  the  required  parts, 
and  then  use  that  part  in  the  computation  of  other  parts.  If, 
however,  he  make  an  error  in  the  first  part,  that  error  is  repeated 
and  perhaps  magnified,  in  the  computation  of  all  other  parts  which 
depend  upon  it  ;  hence  the  importance  of  testing  the  results. 

NOTE  3.  Unless  he  use  the  special  tables  noted  under  the  right 
plane  triangle,  the  computer  must  avoid  angles  near  0°,  90°  or 


86  SPHERICAL   TRIGONOMETRY.  [IV. 

180°.  For  this  purpoca  he  may  prove  and  use  the  following 
formulae  : 

173]  sinHr  =  sinHp  cosHg-h  cos2|-£>  sin2-|-g ;  [153,61,36 
174]  tanr^p  =  tan±(r  +  q)  tan|(r  -  q)  ;  [65,  153,  77 

175]          tair-i-p  =  sin(r  -  q)  :  sin(r  +  g)  ;  [65, 154,  57 

176]  tan  £P  =  sin(r  —  q) :  smp  cosg ;  =  SJf^f;  [65, 154, 151 
177]  sin(p  —  q.)  =  sinp  tau^p  —  singtan^Q;  [176,40,153 
178]  tan2|p  =  tan  J(Q  -f  P  -  90°)  :  tan|(Q  -  p  +  90°)  ; 

[65, 154,  72 
179]         tan2!?-  =  —  cos(p  +  Q)  :  cos(p  -  Q)  .  [65, 153,  58 

§  5.     SOLUTION    OF    QTJADRANTAL    TRIANGLES,    AND 
ISOSCELES    TRIANGLES. 

PROB.  2.    To  SOLVE  A  QUADRANTAL  TRIANGLE  : 

Find  the  triangle  which  is  polar  to  the  given  triangle;  it  is  a 
right  triangle;  solve  it,  and  take  the  supplements  of  the  parts 
thus  found  for  the  corresponding  parts  of  the  given  triangle. 

NOTE  1.  Napier's  rules  apply  directly  to  the  quadrantal 
triangle,  if  the  quadrant  is  ignored,  and  for  the  five  parts  are 
taken  the  two  angles  adjacent  to  the  quadrant  and  the  comple- 
ments of  the  opposite  angle  and  the  two  oblique  sides. 

NOTE  2.  Manifestly,  the  biquadrantal  triangle  cannot  be 
solved  unless  either  the  base  or  the  vertical  angle  is  given  ;  for 
the  remaining  parts,  two  right  angles  and  two  quadrants,  are 
quite  independent  of  these  two. 

PROB.  3.    To  SOLVE  AN  ISOSCELES  TRIANGLE  : 

Draw  an  arc  from  the  vertex  to  the  middle  of  the  base,  thereby 
dividing  the  given  triangle  into  two  equal  right  triangles;  solve 
one  of  these  triangles. 

NOTE.  When  only  the  base  and  the  vertical  angle  are  given, 
there  are  two  triangles,  one  triangle  or  none,  according  as  the 
base  <,  =  or  >  the  vertical  angle.  When  only  the  two  equal  sides 
or  the  two  equal  angles  are  given,  there  is  an  infinite  number  of 
triangles.  Otherwise,  subject  to  the  conditions  in  §  1,  there  is 
one  triangle,  and  but  one. 


§O.J  SOLUTION    OF    SPHERICAL   TRIANGLES.  87 

§  6,     SOLUTION    OF    OBLIQUE    TRIANGLES. 

PROB.  4.    To  SOLVE  AN  OBLIQUE  TRIANGLE. 
FIRST  METHOD.     By  means  of  right  triangles. 


—  —  D 


Let  ABC  be  any  oblique  spherical  triangle,  and  a,  b,  c  the  sides 
opposite  the  vertices  A,  r.,  (5  respectively.  Let  N  be  the  pole  of 
AB,  and  through  N  and  c  draw  a  great  circle,  meeting  the  great 
circle  AB  at  D  and  i>',  whereof  D  stands  less  than  180°  in  the 
direction  AB  from  A  ;  then  is  DC  _L  AB.  [geom. 

Let  p  stand  for  DC  ;  </',  always  positive,  for  AD  ;  q"  for  DB  ; 
c',  always  positive,  for  Z  ACD  ;  c"  for  Z  DCB. 

CASE  1.  Given  two  sides  and  the  included  angle,  for  example 
6,  c  and  A  ;  then  : 

In  rt.  A  ACD,  b  and  A  are  known,  whence  p,  q'  and  c'  are 

found  ; 
q"  =  c-q'', 
In  rt.  A  BCD,  p  and  q"  are  known,  whence  a,  Z  CBD  and 

<•"  are  found  ; 
c  =  c'+c".« 

NOTE  1.  c"  is  positive  or  negative,  and  B  =  CBD  or  180°— CBD, 
according  as  q"  is  positive  or  negative  with  reference  to  AB,  i.e., 
according  as  c  >  or  <  q'. 

NOTE  2.    There  is  always  one  triangle,  and  but  one. 

The  parts  arc  determined  without  ambiguity  by  the  formulae. 


88  SPHERICAL   TRIGONOMETRY.  [IV. 

CASE  2.     Given  two  angles  and  the  included  side,  for  example 
c,  A  and  b  ;  then  : 

In  rt.  A  ACD,  b  and  A  are  known,  whence  p,  q'  and  cf  are 

found ; 
c"  =  c  —  c'; 
In  rt.  A  BCD,  p  and  c"  are  known,  whence  a,  q"  and  Z  CBD 

are  found ; 


NOTE  1 .  q"  is  positive  or  negative,  and  B  =  CBD  or  180°— CBD, 
according  as  c"  is  positive  or  negative  with  reference  to  positive 
rotation,  i.e.,  according  as  c  >  or  <  c'. 

NOTE  2.    There  is  alwa}~s  one  triangle,  and  but  one. 

The  parts  are  determined  without  ambiguity  from  the  formulae. 

NOTE  3.  If  for  the  given  triangle  its  polar  be  substituted, 
then  two  sides  and  the  included  angle  are  known,  and  Case  2  is 
embraced  in  Case  1 .  The  required  parts  are  the  supplements  of 
the  parts  found  in  the  polar. 


CASE  3.     Given  tiuo  sides  and  an  angle  opposite  one  of  them, 
for  example  a,  b  and  A  ;  then  : 

In  rt.  A  ACD,  b  and  A  are  known,  whence  jp,  q'  and  c'  are 

found ; 

In  rt.  A  BCD,  p  and  a  are  known,  whence  q",  Z.  CBD  and 
c"  are  found ; 


§  G.}  SOLUTION   OF   SPHERICAL   TRIANGLES.  89 

NOTE  1.  q'  is  full}'  determined  from  the  data,  but  q"  (found 
from  its  cosine)  ma}*  be  positive  or  negative  ;  and  there  are  two 
triangles,  one  triangle  or  none,  according  as  q'—q'(  and  q'+q", 
both,  one  of  them  or  neither,  lie  between  0°  and  180°. 

The  perpendicular  CD  falls  within  or  without  the  triangle,  and 
B  =  CBD  or  180°—  CBD,  according  as  q"  is  taken  positive  or  nega- 
tive with  reference  to  the  direction  AB. 
And,  v   A  and  Z  CBD  are  always  of  the  same  species  with  j?, 

[Thm.  1  Cor.  I 
.'.  A  and  B  are  of  the  same  species  or  different  species 

according  as  B  =  CBD  or  180°—  CBD  ; 
i.e.,          according  as  q"  is  positive  or  negative. 

*NOTE  2.  "Whether  there  are  two  triangles,  one  triangle  or 
none,  may,  in  general,  be  known  by  inspection  of  the  given  parts  : 

(1)  If  a  lies  between  b  and  its  supplement,  there  is  one  tri- 
angle, and  but  one. 

(2)  If  a  equals  b  or  its  supplement,  then  : 

If  A  is  of  the  same  species  with  a,  and  is  not  90°,  there 

is  one  triangle ; 

If  A  is  of  different  species  from  a,  there  is  no  triangle  ; 
If  A,  a  and  b  are  all  90°,  there  is  an  infinite  number  of 

triangles. 

(3)  If  b  lies  between  a  and  its  supplement,  and  if  a  and  A 
are  of  the  same  species,  there  are  two  triangles,  one  triangle  or 
none,  according  as  a  is  nearer  90°  than  p,  as  near  it,  or  more 
remote  from  it ;    and  there  is  no  triangle  if  a  and  A  are  of 
different  species. 

For  *.•     A  BCD  is  possible  if  a  is  as  near  90°  as  p,  and  of  the 

same  species ; 

.•.     it-is  possible  if  a  is  as  near  90°  as  6,  and  of  the  same 
species.  [Prob.  1,  Case  2,  note 

And  • .  •     cos  a  =  cos  p  cos  q"    and     cos  b  =  cosp  cos  q',          [153 
cos  a  :  cos  b  =  cos  q"  :  cos  q1,  [Thm.  prop'n 

a  proportion  wherein  a,  b  and  q'  are  known,  and  q"  may,  in  gen- 
eral, be  either  positive  or  negative. 


90 


SPHERICAL   TRIGONOMETRY. 


[IV. 


(1) 

then 

and 


(2) 
and 


But 

then 

So, 
then 


If  a  is  nearer  90°  than  5, 

q"  is  nearer  90°  than  q',  [I.   §  23,  Note  2 

of  q'-\-q"  and  q'—q",  one  always,  but  never  both,  lies 

between  0°  and  180° ;  the  first  or  the  second  of  them, 

according  as  q' <  or  >  90°. 
If  a  =  b,    then    q"=q',  [2 


in  either  case,  in  general,  of  q'  -f-  q"  and  q'  —  q",  one 
lies  between  0°  and  180°,  and  the  other  =  0°  or  180°, 
and  there  is  one  triangle. 

if  a  and  A  are  not  of  the  same  species, 

a  differs  from  90°  not  less  than  &, 

there  is  no  triangle.  „  [Thm.  3  Cor. 

if  a,  6  and  A  are  all  90°, 

B  is  also  90°,  and  the  triangle  is  biquadrantal  and  in- 


determinate. 


[Thm.  1  Cor.  3 


\c 


(3) 

then, 

and, 

But, 

then 

So, 
then 


If  b  is  nearer  90°  than  a, 

if  there  is  a  triangle  ACD,'</'  is  nearer  90°  than  q", 

in  general,  q'+q"  and  (/'-(/"both  lie  between  0°and  180°. 

ifp  =  a, 

g"  =  0,    and    q'+q"=  q'-  q", 

there  is  but  one  (a  right)  triangle. 

if  p  is  nearer  90°  than  a, 

cos#'f>  1,  which  is  impossible  ; 

there  is  no  triangle  ACD,  and  no  triangle  ABC. 


And*.*     b  is  nearer  90°  than  a, 


[hypoth. 


there  is  no  triangle  when  a  and  A  are  of  different  species. 

[Thm.  3  Cor. 


§  G.]  SOLUTION   OF    SPHERICAL   TRIANGLES.  91 

CA-I;  -1.  Given  two  angles  and  a  side  opposite  one  of  them, 
for  example  A,  B  and  a. 

In  rt.  A  BCD,  a  and  Z  CBD  are  known,  whence  p,  q"  and 

c"  are  found ; 
In  rt.  A  ACD,-p  and  A  are  known,  whence  6,  q'  and  c'  are 

found ; 
c  =  </  +  <?"; 
c  =  c'  +  c". 

NOTE  1.  c"  is  full}'  determined  by  the  data,  but  c'  (found  from 
its  sine)  may  be  less  or  greater  than  90°  ;  [Prob.  1,  Case  4,  note 
and  there  are  two  triangles,  one  triangle  or  none,  according  as 
c'  —  c"  and  c'  +  c",  both,  one  of  them  or  neither,  lie  between  0° 
and  180°. 

The  perpendicular  CD  falls  within  or  without  the  triangle,  and  c" 
is  positive  or  negative  with  reference  to  positive  rotation,  according 
asp  and  B  are  of  the  same  or  different  species  ;  [Thm.  1  Cor.  1 
and  b  is  less  or  greater  than  90°  according  as  A  and  c'  are  of  the 
same  or  different  species.  [Thm.  1  Cor.  2 

*NOTE  2.  Whether  there  are  two  triangles,  one  triangle,  or 
none,  may,  in  general,  be  known  b}T  inspection  of  the  given  parts  : 

(1)  If  A  lies  between  B  and  its  supplement,  there  is  one  tri- 
angle, and  but  one. 

(2)  If  A  equals  B  or  its  supplement,  then  : 

If  a  is  of  the  same  species  with  A,  and  is  not  90°,  there 

is  one  triangle ; 

If  a  is  of  different  species  from  A,  there  is  no  triangle  ; 
If  a,  A  and  B  are  all  90°,  there  is  an  infinite  number  of 

triangles. 

(3)  If  B  lies  between  A  and  its  supplement,  and  if  A  and  a  are 
of  the  same  species,  there  are  two  triangles,  one  triangle,  or  none, 
according  as  A  is  nearer  90°  than  p,  as  near  it,  or  more  remote 
from  it ;  and  there  is  no  triangle  if  A  and  a  are  of  different  species. 
For  •.*  j  ACD  is  possible  if  A  is  as  near  90°  as  p,  and  of  the 

same  species ; 

.  •.     it  is  possible  if  A  is  as  near  90°  as  B,  and  of  the  same 
species.  [Prob.  1,  Case  4,  note 


92 


SPHERICAL   TRIGONOMETRY. 


[IV. 


And-.-     cos  A  =  cos  j>  sine'     and     cos  CBD  =  cos  p  sine",      [154 
.'.     cos  A  :  cos  CBD  =  sine' :  sine",  [Thm.  prop'n 

a  proportion  wherein  A,  CBD  and  c"  are  known,  and  c'  may,  in  gen- 
eral, have  two  supplementary  values,  c:  and  C2,  both  positive.   [10 


a) 

then 
and 

(2) 
and 


But, 
then 

So, 
then 


If  A  is  nearer  90°  than  B, 

c"  is  nearer  90°  than  cl  and  C2,  [§  23,  Note  2 

of  Cj  -f-  c"  and  c2  +  c",  one  alwaj^s,  but  never  both,  lies 

between  0°  and  180°. 
KA  =  B,    then     d  =  c"   or    c2  =  c", 
if  A  =  180°- B,   then    d  =  -c"   or   c2  =  -c"; 
in  either  case,  in  general,  of  cx  +  c"  and  C2  -f-  c",  one 

lies  between  0°  and  180°,  and  the  other  =  0°or  180°, 

and  there  is  one  triangle, 
if  A  and  a  are  not  of  the  same  species, 
A  differs  from  90°  not  less  than  B, 
there  is  no  triangle.  [Thm.  4  Cor.  1 

if  A,  B  and  a  are  all  90°, 
6  is  also  90°,  and  the  triangle  is  biqnadrantal  and  in- 


determinate. 


[Thin.  1  Cor.  3 


(3) 

then 

and, 

But, 

then 

So, 
then 


If  B  is  nearer  90°  than  A, 

if  there  is  a  triangle  ACD,  cf  is  nearer  90°  than  c", 

in  general,  Cj-f  c"and  c2+c"both  lie  between  0°  and  180°. 


c'  =  90°,    and    d  +  c"  =  c,  -f-c", 
there  is  but  one  (a  qnadrantal)  triangle. 
if  p  is  nearer  90°  than  A, 
sinc'>  1,  which  is  impossible  ; 


§  G.]  SOLUTION  OF  SPHERICAL   TRIANGLES.  93 

.'.  there  is  no  triangle  ACD,  and  no  triangle  ABC. 
And*/  B  is  nearer  90°  than  A,  [hypoth. 

.'.  there  is  no  triangle  when  A  and  a  are  of  different  species. 

[Thm.  4  Cor.  1 

NOTE  3.  If,  for  the  given  triangle,  its  polar  be  substituted, 
then  the  two  sides  and  an  angle  opposite  one  of  them  are  known, 
and  Case  4  is  embraced  in  Case  3.  The  required  parts  are  the 
supplements  of  the  parts  found  in  the  polar. 

CASE  5.     Given  the  three  sides,  a,  fr,  c  ;  then  : 

•/ cosa  =  cos p  cos  q"     and     cos6  =  cosp  cosqf,  [153 

.'.  cos  a  :  cos  b  =  cosq"  :  cos  7',  [Thm.  prop'n 

and      cosa-f-cos6  :  cosa  —  cos 6  =  cos ^"+ cos g' :  cos g"— cos q'. 
But  v  cos  a + cos  6  :  cosa— cosb=  —  cot^(a-f-&)cot-J(a— &)?  [77 
and       cos  q'  +  cos  q"  :  cos  q"  —  cos  q' 

=  -  coti(r/+  q")  coti(f/'-  q') , 
and  •.•c  =  q'+q" ; 

.'.  cot  J-  (a  -}-  b)  cot  %  (a  —b)  =  cot^-c  cot-J-  (q"—  q!) ,  whence 
i  (Y'-r/)  is  found. 

Andvr/=K-W-2')     and     q"=$c  +  £(</"-  q1), 

.'.In  rt.  A  ACD,  b  and  q1  are  known,  whence  A  and  c'  are 

found  ; 
In  rt.  A  BCD,  a  and  q"  are  known,  whence  Z  CBD  and  c" 

are  found  ; 
c  =  c'  +  c".  ' 

NOTE  1.  B  =  CBD  or  180°— CBD,  and  c"  is  positive  or  nega- 
tive with  reference  to  positive  rotation,  according  as  q"  is 
positive  or  negative  with  reference  to  AB. 

NOTE  2.  If  a  +  b  +  c  =  or  >  360°,  or  if  either  side  =  or  >  the 
sum  of  the  other  two,  there  is  no  triangle  ;  otherwise  there  is 
one  triangle,  and  but  one. 

The  parts  are  determined  without  ambiguity  by  the  formulae. 

CASE  6.     Given  the  three  angles,  A,  B,  c  ;  then: 

v  cos  A  =  cosp  sine',    and     cos  B  =  cos  p  sine",  [154 

. ' .  cos  A  :  cos  B  =  sin  c' :  sin  c" ;  [Thm .  prop'n 

and       cos  A+  COSB  :  COSA  —  COSB  =sinc'+  sine"  :  sine'—  sine". 


94  SPHERICAL   TRIGONOMETRY.  [IV. 

Bllt  V  COSA  +  COSB  :  CObA  —  COSB  =  —  COtJ(A+B)cotJ(A  —  B),[77 

and       sinc'+sinc"  :  sine'—  sinc'^=tani(c'H-c'')cot^c'-c'0,[72 
and  vc  =  c'  +  c"  ; 

.*.  —  cot£(A  +  B)  cot^(A—  B)=tan^c  cot  J(c'—  c"),  whence 

i-(c'-  c")  is  found. 
And  v  c'=  ic  +  £(c'-  c")    and    c"=  Jc  -  |(c'-  c")  , 

.'.In  rt.  A  ACD,  A  and  c'  are  known,  whence  b  and  q1  are 

found  ; 

In  rt.  A  BCD,  Z  CBD  and  c"  are  known,  whence  a  and  q" 
are  found  ; 


NOTE  1.  B  =  CBD  or  180°  —  CBD,  and  q"  is  positive  or  negative 
with  reference  to  AB,  according  as  c"  is  positive  or  negative  with 
reference  to  positive  rotation. 

NOTE  2.  If  A  +  B  -f  c  does  not  lie  between  180°  and  540°,  or 
if  either  angle  does  not  exceed  the  difference  between  180°  and 
the  sum  of  the  other  two  angles,  there  is  no  triangle  ;  otherwise 
there  is  one  triangle,  and  but  one. 

The  parts  are  determined  without  ambiguity  by  the  formulae. 

NOTE  3.  If  for  the  given  triangle  its  polar  be  substituted, 
then  the  three  sides  are  known,  and  Case  6  is  embraced  in 
Case  5.  The  required  parts  are  the  supplements  of  the  parts 
found  in  the  polar. 

SECOND  METHOD.     By  means  of  the  general  properties. 

CASE  1.  Given  two  sides  and  the  included  angle  ,  for  example 
6,  c  and  A. 

Find  i(B  +  c)  and  KB  —  c)  bJ  Thm-  12>  theu  : 

B  =  i(B-f-c)-f  !(B-C),  and  C  =  -|-(B  +  c)  —  KB  —  c), 


sina  =      ±  .  sin  A,  whence  a  is  found.  [Thm.  2 

sin  B 

NOTE.     There  is  alwajrs  one  triangle,  and  but  one. 
The  parts  B  and  c  are  determined  without  ambiguity  Uy  the 
formulae,  and  the  species  of  a  is  determined  by  Thm.  12,  Cor. 


§6.J  SOLUTION   OF   SPHERICAL   TRIANGLES.  95 

CASE  2.  Given  two  angles  and  the  .included  side,  for  example 
c,  A  and  b. 

Find  |(c  +  a)  and  £(c  —  a)  by  Thm.  12  ;  then  : 

c  =  J(c  +  a)  +  £(c  —  a)  ,  and  a  =  £(c  +  a)  —  |(c  —  a)  , 

sinB  =  ^^sin6,  whence  B  is  found.  [Thm.  2 

sin  a 

NOTE  1.    There  is  always  one  triangle,  and  but  one. 

The  parts  c  and  a  are  determined  without  ambiguity  by  the 
formulae,  and  the  species  of  b  is  determined  by  Thm.  12,  Cor. 

NOTE  2.  The  solution  may  also  be  obtained  by  applying  the 
methods  of  Case  1  to  the  polar  triangle. 

CASE  3.  Given  two  sides  and  an  angle  opposite  one  of  them, 
for  example  a,  b  and  A  ;  then  : 


shiB  =         sin&,  whence  B  is  found.  [Thm.  2 

sin  a 

Find  c  and  c  from  the  formulae  of  Thin.  12. 

NOTE.     There  may  be  two  triangles,  one  triangle  or  none. 

If  sin  A  sin  b  <  sin  a,  in  general,  there  are  two  triangles  ; 
for  then  sinB<  1,  and  B  (determined  from  its  sine)  may  be  either 
of  two  angles  which  are  supplementary  to  each  other,  and  the 
side  CB  may  lie  to  the  right  or  to  the  left  of  CD. 

But  this  is  limited  by  the  condition  that  the  greater  angle  lies 
opposite  the  greater  side,  and  that  no  angle  or  side  can  exceed 
180°,  or  be  negative. 

If  sin  A  sin  b  =  sin  a,  there  is  one  (a  right)  triangle  ; 
for  then  sinB  =  1,  and  B  is  a  right  angle. 

If  sin  A  sin  b  >  sin  a,  there  is  no  triangle  ; 
for  then  sinB  >  1,  which  is  impossible. 

For  detail  of  specific  conditions  the  reader  may  consult  the 
note  to  First  Method  for  solving  this  case. 

CASE  4.  Given  two  angles  and  a  side  opposite  one  of  them, 
for  example  A,  B  and  a  ;  then  : 

ain  b  =  !1M.  sinB,  whence  b  is  found.  [Thm.  2 

sinA 

Find  c  and  c  from  the  formulae  of  Thm.  12. 


96  SPHERICAL  TRIGONOMETRY.  [IV. 

NOTE  1.     There  ma,y  be  two  triangles,  one  triangle  or  none. 

If  sin  a  sin  B  <  sin  A,  in  general,  there  are  two  triangles  ; 
for  then  sin  6  <  1 ,  and  b  (determined  from  its  sine)  may  be  either 
of  two  arcs  which  are  supplementary  to  each  other. 

But  this  is  limited  by  the  condition  that  the  greater  side  lies 
opposite  the  greater  angle,  and  that  no  side  or  angle  can  exceed 
180°,  or  be  negative. 

If  sin  a  sinB  =  sin  A,  there  is  one  (a  quadrantal)  triangle  ; 
for  then  sin&  =  1,  and  b  is  a  quadrant. 

If  sin  a  sinB  >  sin  A,  there  is  no  triangle  ; 
for  then  sin  b  >  1,  which  is  impossible. 

For  detail  of  specific  conditions  the  reader  may  consult  the 
note  to  First  Method  for  solving  this  case. 

NOTE  2.  The  solution  may  also  be  obtained  by  applying  the 
methods  of  Case  3  to  the  polar  triangle. 

CASE  5.     Given  the  three  sides,  a,  &,  c. 
Apply  the  formulae  of  Thm.  5,  6  or  7. 

NOTE  1 .  If  either  side  equals  or  exceeds  the  sum  of  the  other 
two,  or  if  the  sum  of  the  three  sides  equals  or  exceeds  360°,  there 
is  no  triangle  ;  otherwise  there  is  one  triangle,  and  but  one. 

That  there  is  a  single  triangle  appears  also  from  the  formulae, 
since  the  half-angles  computed  must  be  each  less  than  90°,  and 
but  one  such  half-angle  can  be  found  from  a  given  function. 

NOTE  2.  Of  these  formulae  those  of  Thm.  5  use  nine  different 
logarithms,  those  of  Thm.  6  use  ten  different  logarithms,  and 
those  of  Thm.  7  use  only  seven  different  logarithms  for  the  com- 
putation of  all  the  angles.  Those  of  Thm.  7  are,  therefore,  gen- 
erally to  be  preferred  ;  they  may  be  put  in  the  form 

!  1  /  sin  (s  —  a)  sin  (3  —  6)  sin(s  —  c) 

tan  7T  A  — »*  I  — ; 9 

sm(s  — a)\  sins 

1  I  sin  (s  —  a)  sin  (s  —  b)  sin(s  —  c) 

tan  -s-  B  =  — — — -  ^  I  • ; ? 

sm(s  —  b)  \  sins 

^    __ 1_         |  sin  (s  —  a}  sin  (s  —  b)  sin  (s  —  c) 

~~  sin  (s  —  c)  V  sin  s 


§6.]  SOLUTION  OF   SPHERICAL  TRIANGLES.  97 

wherein  the  second  factor  of  the  right  member  is  the  same,  and 
may  be  computed  once  for  all. 

NOTE  3.  A  complete  check  is  afforded  by  either  of  Delambre's 
formulae,  [165-168]. 

CASE  6.    Given  the  three  angles  A,  B,  c. 

Apply  the  formulae  of  Thm.  8,  9  or  10. 

NOTE  1 .  If  either  angle  does  not  exceed  the  difference  between 
180°  and  the  sum  of  the  other  two,  or  if  the  sum  of  the  angles 
does  not  lie  between  180°  and  540°,  there  is  no  triangle  ;  other- 
wise, there  is  one  triangle,  and  but  one. 

That  there  is  a  single  solution  appears  also  from  the  formulae, 
since  the  half-sides  computed  must  be  less  than  a  quadrant,  and 
but  one  such  half-side  can  be  found  from  a  given  function. 

NOTE  2.  Among  the  formulae  the  same  choice  may  be  made 
as  in  Case  1.  The  reader  may  transform  those  of  Thm.  10  for 
convenient  use. 

NOTE  3.  A  complete  check  is  afforded  by  either  of  Delambre's 
formulae,  [165-168]. 

NOTE  4.  The  solution  may  also  be  obtained  by  apptying  the 
methods  of  Case  5  to  the  polar  triangle. 


The  following  formulae,  which  the  reader  may  deduce  from 
[159-164],  are  sometimes  of  use  : 

180"1     s*n  s          _  cos-J-Acos^-B      cos  s  _sin^-#sin|-&  m 

sine  sin^-c  sine  cos^-c 

sin  (s  —  c)  _  sin  j-  A  sin  -^  B      cos(s  —  c)  __  COS^CT 


sine  sin^c  sine  cos^c 

sin(s  —  d)  _  cos  j-A  sin  j-B  cos(s  —  A)  _  sin  j-q  cos-^-6 

sine  cos^-c  sine  sin^-c 

183"!     sm(g  —  c)_tanj-A  cos(s  —  c)  _ 


sin(s—  a)      tan^-c  cos(s  —  A)      cot-^-c 

-  A>  =- 


184] 

sm  s  cos  s 

185]     sin(s—  a)tan  ^A=sin(s—  6)tan^B=sin(s—  c)tanjc.  [183 


98  SPHERICAL   TRIGONOMETRY.  [IV. 

§  7,     RELATIONS   BETWEEN   SPHERICAL   AND    PLANE 
TRIGONOMETRY. 

Manifestly,  when  the  sides  of  a  spherical  triangle  subtend  very 
small  angles  at  the  center  of  the  sphere,  the  spherical  triangle 
differs  but  little  from  a  plane  triangle  having  the  same  vertices ; 
and,  if  the  vertices  be  fixed  in  position  while  the  center  of  the 
sphere  recedes  further  and  further  away,  and  the  radii  grow 
longer  and  longer,  then  the  spherical  triangle  approaches  closer 
and  closer  to  the  plane  triangle  having  the  same  vertices.  Then 
also  the  small  angles  at  the  center  of  the  sphere,  which  are 
measured  by  the  sides  of  the  triangle,  are  very  nearly  equivalent 
to  their  sines  or  tangents,  and  the  sum  of  the  three  angles  of  the 
triangle  is  very  nearly  180°. 

If,  in  the  several  formulae  of  spherical  trigonometiy,  a  is  sub- 
stituted for  sin  a,  tana,  2  sin^-a, ;  p,  for  ship, ;  s,  for  sins, 

;  and  1,  for  cos  a, ;  then,  in  general,  these  formulae  reduce 

to  the  corresponding  formulae  of  plane  trigonometry,  or  to  mere 
truisms.     Thus : 

Spherical.  Plane. 

151]  snrp  =  sin?*  sin  P=  tang  cot  Q  ;  p  =  r  sinp  =  gcotQ 

or    y  =  r  sin  o  =  x cotp.    [I.  §  7 
154]  cos  P = cosjp  sin  Q= tang  cot  r;    cosp=l  .  sinQ=g  :  r 

or     coso=sinp  =  o':  r.     [I.  §  7 
156]  sina:  sin6  =  sinA:  sins  ;  a  :  b  =shiA  :  sine.        [103 

IKO-I     •    i  I  sin  (8 -6)  sin  (g-c)  . 

1591  smlA=A  : — 7—- » 

\  sin  6  sine 

i  l\^        )  \^ ~~ ^/      nno 

rfB.jA.-y-  -w- 

f sins  sin  (s  —  a)  , 
160]  COS|A  =  ^       Sin6sinc 

'   >-'•) 


§  7.]  SOLUTION  OF   SPHERICAL   TRIANGLES.  99 

Spherical.  Plane. 

166]  gtoKA~B)  =  "in*.(g1~6)co«io; 
sm^c 

smJ(A~B)  =  ^^cos|c.    [121 


168] 

OAll-R-ly 

•f    s  \  a  "y™  0      •        i  r--|  ^\r» 

cir»  ?L(  ft.  r^j  7A 

170] 


sm  -    a 

~B)==^^cotic.    [107 
a  -j—  o 


c.     [121 

*  Some  correspondences  between  the  formulae  of  plane  and  of 
spherical  trigonometry  appear  only  when  functions  of  the  sides, 
of  their  half-sum,  etc.,  are  represented  by  taking  two  or  more 
terms  of  the  series  [94-99]  instead  of  using  the  first  term  alone 
as  above.  Thus  : 

Formula  cosr  =  cos_p  cos  q  [153 

becomes  a  mere  truism  if  each  cosine  be  represented  by  1  ,  its 
limit  when  j>,  q  and  r  become  indefinitely  small  ;  but  if  the 
values 


cos?*  =    — 

[95 
be  substituted  in  [153]  they  give 


100  SPHERICAL  TEIGONOMETKY.  [IV. 

.•.  p2  -}-  (f  =  j*2  ±  terms  of  higher  degree,  whose  ratios  to 
j92,  (f  and  r2  have  the  limit  0  when  p,  q  and  r  become 
indefinite!}7  small ; 

i.e.,          p2  +  #2,  or  x2  4-  ?/2,  =  j*2,  whence  sin20  +  cos20  =  1 ; 
.-.     [153]  corresponds  to  [36]. 

So,  writing  b  for  sin6,  c  for  sine,  1  —  Ja2+ for  cosa, , 

then  the  formula 

cosa  =  cos  6  cose -J- sin  6  sine  cos  A  [157 

gives        a2  =  b2  +  c2  -  2  be  cos  A.  [108 

So,  writing  two  terms  of  [94]  for  sin6, ,  and  three  terms 

of  [95]  for  cos  a, ,  then  the  formula 

cos  a  =  cos  b  cose  4-  sin  b  sine  cos  A  [157 

gives        &C(COSA'— COSA)  =  1J2-(a262+&2c2+c2a2)  — 2^(a4+64+c4) 

±  terms  whose  limiting  ratios  to  these  terms,  when 

a,  b  and  c  become  indefinitely  small,  are  0  ; 

and  so  for  ca(cosBf—  COSB),  and  a&(cosc'—  cose) .    [symmetry 

.*.     be  (cos  A'  —  cos  A)  =  ca  (COSB '—COSB)  =ab  (cose  '—cose) 

approximately,  wherein  cos  A'=  (62+c2— a2) :  2  6c, , 

and  A',  B'  and  c'  are  the  angles  of  the  plane  triangle  whose 
sides  a',  6'  and  c'  are  respectively  equal  to  the  arcs 
a,  b  and  c. 

But  '. '     cos  A'—  COSA=  2sin|-(A  —  A')sin-J(A-f- A') 
=  (A  —  A')  sin  A',  very  nearly, 
and  so  for  COSB'— COSB     and    cose'— cose,  [symmetry 

&C(A  —  Af)sinAf=  C«(B  —  B')sinBf=a&(c  —  c')sincf, 
.-.     A  —  A'  =  B  —  B'  =  C  —  c',  very  nearly ;  [156 

186].'.  each  angle  of  sph. AABC  exceeds  the  corresponding 
angle  of  pi.  AA'B'C'  by  one-third  of  the  spherical  ex- 
cess, A  +  B  -f-  c  —  180°;  which  is  Legendre's  theorem. 

§  8,     EXEKCISES. 

Solve  the  spherical  right  triangles,      given  : 

1.  p=116°,  g=    16°,  R  =  90°. 

2.  r=140°,  p=    20°,  R  =  90°. 

3.  P=    80°  10',  g=155°  46',  R  =  90°. 


§8.]  SOLUTION   OF   SPHERICAL  TRIANGLES.  101 

4.  p=100°,  j9  =  112°,  R  =  90°. 

5.  r=120°,  p=120°,  R  =  90°. 

6.  p=    60°  47',  Q=    57°  16',  R=90°. 

7.  r=140°,  p  =  140°,  R=90°. 

8.  r  =  120°,  p=    90°,  R  =  90°. 
Solve  the  quadrantal  triangles,      given  : 

9.  A  =  80°,  a=   90°,  6  =  37°. 

10.  B  =  50°,  6  =  130°,  c  =  90°. 
Solve  the  isosceles  triangles,      given  : 

11.  a=    70°,  6=    70°,  A  =  30°. 

12.  a  =    30°,  A  =    70°,  B  =  70°. 

13.  a  =119°,  6  =  119°,  c  =  85°. 

Solve  the  oblique  triangles  [both  methods] ,      given  : 

14.  6=    98°  12',  c=    80°  35',  A=    10°  16'. 

15.  A  =135°  15',  c=    50°  30',  6=    69°  34'. 

16.  a=   40°  16',  6=   47°  14',  A=    52°  30'. 

17.  a  =  120°,  6=    70°,  A  =130°. 

18.  a=   40°,  6=    50°,  -A  =    50°. 

19.  A  =132°  16',  B  =  139°44',  a  =127°  30'. 

20.  A  =  110°,  B=    60°,  a=    50°. 

21.  A=    70°,  B  =  120°,  a=   80°. 

22.  a  =100°,  6=    50°,  c=    60°. 

23.  A  =  120°,  B  =  130°,  c=    80°. 

24.  In  astronomy  the  altitude  of  a  heavenly  body  is  its  angular 
elevation  above  the  horizon,  and  its  azimuth  is  its  angular  dis- 
tance west  from  the  south  point.     What  is  the  angular  distance 
between  the  moon,  alt.  40°,  az.  25°  w.,  and  Venus,  alt.  24°, 
az.  110   w.  ? 

25.  In  navigation  the  shortest  distance  from  port  to  port 
is  the  arc  of  a  great  circle.     Find  the  course  and  distance  from 
San  Francisco,  lat.  37°  48'  N.,  long.  122°  25'  w.,  to   Cape  of 
Good  Hope,  lat.  33°  56'  s.,  long.  18°  29'  E.,  no  allowance  being 
made  for  intervening  lands. 


102  SPHERICAL   TRIGONOMETRY.  [IV. 

26.  In  geodetic  surveys  triangles  upon  the  earth's  surface  are 
considered  as  spherical  triangles.     Assume  the  earth's  radius  to 
be  3956  miles  ;  then,  if  one  side  of  a  triangle  be  100  miles,  and 
the  adjacent  angles  be  65°  and  60°  respectively  :  Find  the  other 
two  sides  in  degrees  and  in  miles  ;   find  the  third  angle ;  find 
the  spherical  excess  ;  \_geom. 
find  the  area  of  the  triangle  in  square  miles  ;  find  the  number  of 
square  miles  of  area  which  corresponds  to  1"  of  spherical  excess. 

27.  In  a  geodetic  survey  there  were  measured  Z  A  =  30°, 
Z  B  =  48°  45',  Z  c  =  101°  15'  12"  and  side  c  =  70  miles  :  Find  the 
angles  of  the  plane  triangle  whose  sides  equal  a,  b  and  c  of  the 
spherical  triangle,  and  thence  find  the  lengths  of  a  and  b. 

[Legendre's  theorem 

28.  From  the  two  equations 

cos  b  =  cos  c  cos  a  +  sin  c  sin  a  cos  B 

cos  c  =  cos  a  cos  b  -\-  sin  a  sin  b  cose  [157 

eliminate  cose,  and  prove  that 

sin  a  cos  &  =  cos  a  sin  &  cose  +  sine  COSB  ;  [36 

thence  eliminate  sine,  and  prove  that 

sin  a  cot  b  =  cos  a  cose  +  cot  B  sine.  [156 

So,  prove  that 

sin  b  cot  c  =  cos  b  cos  A  +  cot  c  sin  A, 

sine  cot  a  =  cose  COSB  -f  cot  A  sinB, 

sin  a  cote  =  cos  a  COSB  +  cote  sinB, 

sin  b  cota=  cos  b  cose  +cotA  sine, 

sine  cot&  =  cosccosA  +  cotB  sin  A. 

29.  Show  that  [173]  reduces,  for  a  plane  right  triangle,  to 
a^-f-2/2  =  r2;  [174],  to  the  same;    [175],  to  [114];   [176],  to 
[114];  [177], toy— a=ytan£o— ictanj-p;  [178],too-f-p=90°; 
[179],  to  the  same. 

30.  Show  that  [180]  reduces,  for  a  plane  triangle,  to  [122] 
and  s  =  90°  ;  [181],  to  [123]  and  s  =  90°  ;  [182],  to  [125]  and 
a :  c  =  sinA :  sine  ;    [183],  to  (s  —  c)  :  (s  —  a)  =  tan^-A :  tan  Jc 
and  a :  c  =  as  above  ;  [184],  to  what? 


8.]  SOLUTION   OF   SPHERICAL  TRIANGLES.  103 


NOTE  1.  In  this  corollary  ^/(l+sin#)  is  positive  for  all 
values  of  6  from  —  -J-TT  to  -f-j  TT,  negative  for  all  values  of  0  from 
-f-f7r  to  +£TT,  and  so  on;  and  ^/(l—  sin#)  is  positive  for  all 
values  of  9  from  —  f  ?r  to  -f  JTT,  negative  for  all  values  of  6  from 
-f-^TT  to  +J  TT,  and  so  on.  The  reader  may  prove  this  by  observ- 
ing that  when  6  lies  between  0  and  ^TT,  then  cos^-0>smJ0>0  ; 
that  when  0  increases  or  decreases  continuously,  the  radicals 
A/(l-f-sin#)  and  A/(l—  sin#)  can  change  from  positive  to  nega- 
tive or  from  negative  to  positive  only  when  passing  through  0  ; 
i.e.  V(l  +  sin0),  only  when  0  =  —  ITT,  —  JTT,  f?r,  JTT,  •••,  and 
V(l—  sin0),  only  when  0  =  —  |TT,  £TT,  ITT,  ITT,  •••  ;  and  that, 
when  0  passes  through  either  of  these  values,  the  radical 
-N/(l±sin0),  =sin£0±cos!-0,  which  becomes  0,  must  change 
its  sign,  since  both  sin^0  and  cos£0  continue  to  increase  or 
decrease  just  afterward  as  they  did  just  before. 


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